Timer: Off
Random: Off
301. Which expression is used to calculate the escape speed from the surface of a planet of mass \(M\) and radius \(R\)?
ⓐ. \(v_e = \sqrt{\frac{GM}{R}}\)
ⓑ. \(v_e = \sqrt{\frac{2GM}{R}}\)
ⓒ. \(v_e = \sqrt{\frac{GM}{R^2}}\)
ⓓ. \(v_e = \sqrt{\frac{2GM}{R^2}}\)
Correct Answer: \(v_e = \sqrt{\frac{2GM}{R}}\)
Explanation: Escape speed is derived by equating kinetic energy with gravitational potential energy: \(\tfrac{1}{2}mv^2 = \tfrac{GMm}{R}\). Solving gives \(v_e = \sqrt{\tfrac{2GM}{R}}\).
302. If Earth’s mass is \(6 \times 10^{24} \, kg\) and radius is \(6.4 \times 10^6 \, m\), what is Earth’s escape speed? (\(G = 6.67 \times 10^{-11}\))
ⓐ. \(7.9 \, km/s\)
ⓑ. \(9.8 \, km/s\)
ⓒ. \(11.2 \, km/s\)
ⓓ. \(12.5 \, km/s\)
Correct Answer: \(11.2 \, km/s\)
Explanation: Substituting values into \(v_e = \sqrt{\tfrac{2GM}{R}}\), we get \(v_e \approx 11.2 \, km/s\).
303. If the radius of a planet doubles while its mass remains constant, what happens to escape speed?
ⓐ. It becomes double
ⓑ. It becomes half
ⓒ. It decreases by factor \(\sqrt{2}\)
ⓓ. It remains same
Correct Answer: It decreases by factor \(\sqrt{2}\)
Explanation: Since \(v_e \propto \sqrt{\tfrac{1}{R}}\), doubling \(R\) reduces escape velocity by \(1/\sqrt{2}\).
304. For a planet of radius \(R\) and density \(\rho\), escape speed can be expressed as:
ⓐ. \(v_e = \sqrt{\frac{8}{3}\pi G \rho R^2}\)
ⓑ. \(v_e = \sqrt{\frac{GM}{R}}\)
ⓒ. \(v_e = \sqrt{2gR}\)
ⓓ. Both A and C
Correct Answer: Both A and C
Explanation: Using \(M = \frac{4}{3}\pi R^3 \rho\), \(v_e = \sqrt{\frac{2GM}{R}} = \sqrt{\frac{8}{3}\pi G \rho R^2}\). Also, using \(g = \tfrac{GM}{R^2}\), we get \(v_e = \sqrt{2gR}\).
305. Why does escape speed not depend on the direction of projection?
ⓐ. Because gravity is uniform
ⓑ. Because it depends only on gravitational potential energy
ⓒ. Because velocity cancels in all directions
ⓓ. Because it depends only on kinetic energy
Correct Answer: Because it depends only on gravitational potential energy
Explanation: Escape speed is calculated from energy conservation, which is scalar and independent of direction. Thus, only magnitude matters, not direction.
306. What is the escape speed from the Moon’s surface? (Moon’s mass \(7.3 \times 10^{22} \, kg\), radius \(1.74 \times 10^6 \, m\))
ⓐ. \(2.38 \, km/s\)
ⓑ. \(4.9 \, km/s\)
ⓒ. \(7.9 \, km/s\)
ⓓ. \(11.2 \, km/s\)
Correct Answer: \(2.38 \, km/s\)
Explanation: Substituting values into \(v_e = \sqrt{\tfrac{2GM}{R}}\) gives approximately \(2.38 \, km/s\).
307. Escape speed on Mars is about \(5 \, km/s\). What does this imply for spacecraft launches compared to Earth?
ⓐ. They need less fuel to escape Mars than Earth
ⓑ. They need more fuel to escape Mars than Earth
ⓒ. Fuel requirement is same as Earth
ⓓ. Escape is impossible from Mars
Correct Answer: They need less fuel to escape Mars than Earth
Explanation: Since escape speed is proportional to fuel energy required, lower escape speed means less energy and fuel are needed on Mars.
308. Why can Earth’s atmosphere not extend indefinitely into space?
ⓐ. Because gravity vanishes at a certain height
ⓑ. Because atmospheric molecules can achieve escape speed
ⓒ. Because pressure becomes zero after some distance
ⓓ. Because Earth rotates
Correct Answer: Because atmospheric molecules can achieve escape speed
Explanation: Gas molecules with speeds greater than escape velocity can leave Earth’s atmosphere, preventing indefinite extension.
309. If escape speed from Earth is \(11.2 \, km/s\), what would it be for a planet with twice Earth’s radius and same mass?
ⓐ. \(5.6 \, km/s\)
ⓑ. \(7.9 \, km/s\)
ⓒ. \(8.0 \, km/s\)
ⓓ. \(11.2 \, km/s\)
Correct Answer: \(5.6 \, km/s\)
Explanation: Escape speed \(v_e \propto \tfrac{1}{\sqrt{R}}\). Doubling \(R\) reduces escape speed by \(1/\sqrt{2}\), giving about half the Earth’s value.
310. How does escape speed relate to orbital speed at the same radius?
ⓐ. Escape speed is half of orbital speed
ⓑ. Escape speed is \(\sqrt{2}\) times orbital speed
ⓒ. Escape speed is twice orbital speed
ⓓ. They are equal
Correct Answer: Escape speed is \(\sqrt{2}\) times orbital speed
Explanation: Orbital speed at radius \(r\) is \(v_o = \sqrt{\tfrac{GM}{r}}\). Escape speed is \(v_e = \sqrt{2GM/r} = \sqrt{2} \, v_o\).
311. Escape speed is derived by equating which two forms of energy?
ⓐ. Kinetic energy and momentum
ⓑ. Kinetic energy and gravitational potential energy
ⓒ. Potential energy and thermal energy
ⓓ. Rotational energy and kinetic energy
Correct Answer: Kinetic energy and gravitational potential energy
Explanation: Escape speed is obtained by equating \(\tfrac{1}{2}mv^2\) with gravitational potential energy \(\tfrac{GMm}{R}\). This ensures the object just escapes with zero velocity at infinity.
312. What is the gravitational potential energy of a body of mass \(m\) at Earth’s surface?
ⓐ. \(U = -\frac{GMm}{R}\)
ⓑ. \(U = \frac{GMm}{R}\)
ⓒ. \(U = -\frac{GMm}{R^2}\)
ⓓ. \(U = 0\)
Correct Answer: \(U = -\frac{GMm}{R}\)
Explanation: Gravitational potential energy at a distance \(r\) from Earth’s center is \(U = -\frac{GMm}{r}\). At Earth’s surface, \(r = R\).
313. When a body attains escape speed, its total energy becomes:
ⓐ. Negative
ⓑ. Zero
ⓒ. Positive
ⓓ. Infinite
Correct Answer: Zero
Explanation: At escape speed, kinetic energy just balances the magnitude of potential energy. Hence, \(KE + PE = 0\).
314. Why is gravitational potential energy taken as negative?
ⓐ. To represent bound systems under attractive forces
ⓑ. To make equations simpler
ⓒ. To keep escape velocity constant
ⓓ. To cancel out kinetic energy
Correct Answer: To represent bound systems under attractive forces
Explanation: Potential energy is defined zero at infinity. Closer to Earth, energy decreases, making it negative, indicating a bound system.
315. Which condition must be satisfied for a body to escape Earth’s gravitational field?
ⓐ. \(KE = PE\)
ⓑ. \(KE > |PE|\)
ⓒ. \(KE < |PE|\)
ⓓ. \(KE = 0\)
Correct Answer: \(KE > |PE|\)
Explanation: For escape, kinetic energy must be at least equal to the magnitude of potential energy. At exact escape velocity, \(KE = |PE|\).
316. If an object is projected with velocity less than escape speed, what will be its total mechanical energy?
ⓐ. Zero
ⓑ. Positive
ⓒ. Negative
ⓓ. Infinite
Correct Answer: Negative
Explanation: For \(v < v_e\), kinetic energy is insufficient to cancel potential energy, so total energy remains negative. The body remains gravitationally bound to Earth.
317. At infinity, the gravitational potential energy of an object is taken as:
ⓐ. Zero
ⓑ. Positive
ⓒ. Negative
ⓓ. Equal to kinetic energy
Correct Answer: Zero
Explanation: By convention, potential energy at infinity is zero. As the object moves towards Earth, its potential decreases and becomes negative.
318. Escape velocity condition comes from:
ⓐ. \(\tfrac{1}{2}mv^2 = GMm/R\)
ⓑ. \(\tfrac{1}{2}mv^2 = -GMm/R\)
ⓒ. \(\tfrac{1}{2}mv^2 = 0\)
ⓓ. \(\tfrac{1}{2}mv^2 = -U\)
Correct Answer: \(\tfrac{1}{2}mv^2 = GMm/R\)
Explanation: The minimum kinetic energy required to reach infinity equals the magnitude of gravitational potential energy at Earth’s surface.
319. What happens to escape speed if potential energy is doubled (keeping radius constant)?
ⓐ. Escape speed doubles
ⓑ. Escape speed becomes half
ⓒ. Escape speed increases by \(\sqrt{2}\)
ⓓ. Escape speed remains unchanged
Correct Answer: Escape speed increases by \(\sqrt{2}\)
Explanation: Escape speed \(v_e = \sqrt{2GM/R}\). Since \(U \propto M\), doubling \(U\) means doubling \(M\), so \(v_e\) increases by \(\sqrt{2}\).
320. What is the physical significance of escape speed in terms of gravitational potential energy?
ⓐ. It is the velocity at which potential energy becomes maximum
ⓑ. It is the velocity at which kinetic energy equals the magnitude of gravitational potential energy
ⓒ. It is the velocity at which potential energy becomes zero
ⓓ. It is the velocity at which force of gravity vanishes
Correct Answer: It is the velocity at which kinetic energy equals the magnitude of gravitational potential energy
Explanation: Escape speed ensures that \(KE = |PE|\), so total energy becomes zero, allowing the body to escape gravitational influence completely.
321. Why is the escape speed from the Moon much lower than from Earth?
ⓐ. Because Moon rotates faster
ⓑ. Because Moon has less mass and smaller gravity
ⓒ. Because Moon has thicker atmosphere
ⓓ. Because Moon is farther from the Sun
Correct Answer: Because Moon has less mass and smaller gravity
Explanation: Escape speed depends on \(v_e = \sqrt{\tfrac{2GM}{R}}\). The Moon’s much smaller mass compared to Earth results in a lower escape speed (\~2.38 km/s), allowing spacecraft to leave more easily.
322. Why does Earth retain its atmosphere while the Moon does not?
ⓐ. Earth’s escape speed is much higher than average molecular speeds of gases
ⓑ. Moon has no gravitational force
ⓒ. Earth rotates slower than Moon
ⓓ. Moon has too much solar radiation
Correct Answer: Earth’s escape speed is much higher than average molecular speeds of gases
Explanation: Gas molecules on the Moon can easily reach speeds comparable to escape velocity, so gases escape. On Earth, escape speed (11.2 km/s) is far higher than molecular speeds, so atmosphere is retained.
323. Why do rockets require such high energy to leave Earth?
ⓐ. Because Earth’s escape speed is very high (11.2 km/s)
ⓑ. Because atmosphere blocks their path
ⓒ. Because Earth’s rotation is too slow
ⓓ. Because space has no air resistance
Correct Answer: Because Earth’s escape speed is very high (11.2 km/s)
Explanation: Rockets must reach or exceed escape speed to overcome gravity. This requires huge amounts of fuel and energy.
324. Which planet requires the greatest energy to launch a spacecraft?
ⓐ. Mercury
ⓑ. Mars
ⓒ. Jupiter
ⓓ. Venus
Correct Answer: Jupiter
Explanation: Jupiter has very large mass and radius, giving it the highest escape speed (\~59.5 km/s). Thus, escaping its gravitational field requires the most energy.
325. Why can hot hydrogen escape from Earth’s upper atmosphere while oxygen and nitrogen remain bound?
ⓐ. Hydrogen molecules have higher average speeds due to lower mass
ⓑ. Hydrogen is not affected by gravity
ⓒ. Nitrogen and oxygen are heavier due to atmosphere
ⓓ. Oxygen and nitrogen move slower because of sunlight
Correct Answer: Hydrogen molecules have higher average speeds due to lower mass
Explanation: Lighter molecules like hydrogen move faster at the same temperature. Some reach speeds near escape velocity, escaping Earth’s atmosphere, unlike heavier gases.
326. How does escape speed affect the possibility of human colonization on other planets?
ⓐ. High escape speed makes rocket launches more costly and difficult
ⓑ. Low escape speed prevents building houses
ⓒ. Escape speed controls day length
ⓓ. Escape speed controls oxygen availability
Correct Answer: High escape speed makes rocket launches more costly and difficult
Explanation: On planets with high escape velocity (like Jupiter), launching spacecraft would require extreme energy, making colonization impractical.
327. What is the connection between escape speed and black holes?
ⓐ. Escape speed equals the speed of sound near black holes
ⓑ. Escape speed is greater than the speed of light at the event horizon
ⓒ. Escape speed equals orbital speed at infinity
ⓓ. Black holes have no escape speed
Correct Answer: Escape speed is greater than the speed of light at the event horizon
Explanation: A black hole’s escape velocity at its event horizon exceeds \(c\), the speed of light, meaning nothing can escape from it.
328. Why is the escape speed from Mars lower than from Earth?
ⓐ. Mars is closer to the Sun
ⓑ. Mars has smaller mass and weaker gravity
ⓒ. Mars rotates faster than Earth
ⓓ. Mars has no atmosphere
Correct Answer: Mars has smaller mass and weaker gravity
Explanation: Mars’ smaller mass leads to weaker gravitational pull and lower escape speed (\~5 km/s), requiring less fuel for spacecraft to leave.
329. Which of the following is a practical application of escape speed in space missions?
ⓐ. Calculating the orbital velocity of satellites
ⓑ. Determining the minimum fuel required for interplanetary travel
ⓒ. Measuring surface gravity directly
ⓓ. Estimating day-night cycles on planets
Correct Answer: Determining the minimum fuel required for interplanetary travel
Explanation: Escape speed helps engineers calculate how much velocity a spacecraft must achieve to break free from a planet’s gravity and travel into deep space.
330. Why is escape speed important in understanding planetary atmospheres?
ⓐ. It determines the length of a planet’s year
ⓑ. It helps explain why light gases may escape a planet while heavier ones remain
ⓒ. It controls temperature of the planet
ⓓ. It fixes the radius of the planet
Correct Answer: It helps explain why light gases may escape a planet while heavier ones remain
Explanation: If the average molecular speed of a gas is close to or higher than escape velocity, that gas will escape into space. This explains why small planets often lack dense atmospheres.
331. Which of the following is Earth’s only natural satellite?
ⓐ. Phobos
ⓑ. Deimos
ⓒ. The Moon
ⓓ. Titan
Correct Answer: The Moon
Explanation: The Moon is Earth’s only natural satellite. Phobos and Deimos are Mars’ moons, while Titan orbits Saturn. Artificial satellites are man-made, unlike the Moon.
332. Which of the following is an artificial satellite?
ⓐ. The Moon
ⓑ. Sputnik 1
ⓒ. Phobos
ⓓ. Ganymede
Correct Answer: Sputnik 1
Explanation: Sputnik 1, launched by the USSR in 1957, was the first artificial satellite. Natural satellites like Phobos or Ganymede are celestial bodies orbiting planets.
333. What distinguishes artificial satellites from natural satellites?
ⓐ. Artificial satellites are always larger
ⓑ. Natural satellites orbit stars, not planets
ⓒ. Artificial satellites are man-made and launched into orbit
ⓓ. Natural satellites do not follow orbital paths
Correct Answer: Artificial satellites are man-made and launched into orbit
Explanation: Natural satellites are naturally occurring (like the Moon), while artificial satellites are human-engineered objects placed in orbit for communication, research, or navigation.
334. Which of the following is an example of a communication satellite?
ⓐ. Sputnik 1
ⓑ. INSAT
ⓒ. Voyager 1
ⓓ. Hubble Space Telescope
Correct Answer: INSAT
Explanation: INSAT is an Indian communication satellite series used for broadcasting, telecommunication, and meteorology. Sputnik 1 was experimental, while Hubble is an observatory.
335. Which of the following is an Earth observation satellite?
ⓐ. Cartosat
ⓑ. Aryabhata
ⓒ. Voyager 2
ⓓ. Chandra X-ray Observatory
Correct Answer: Cartosat
Explanation: Cartosat is an Indian Earth observation satellite series used for remote sensing and cartography. Voyager 2 is an interplanetary probe.
336. What type of satellite is the Hubble Space Telescope?
ⓐ. Communication satellite
ⓑ. Remote sensing satellite
ⓒ. Astronomical observatory satellite
ⓓ. Navigation satellite
Correct Answer: Astronomical observatory satellite
Explanation: Hubble is an artificial satellite designed as a space-based observatory to study astronomical phenomena outside Earth’s atmosphere.
337. Which of the following is NOT a purpose of artificial satellites?
ⓐ. Weather forecasting
ⓑ. GPS navigation
ⓒ. Military surveillance
ⓓ. Generating Earth’s gravity
Correct Answer: Generating Earth’s gravity
Explanation: Earth’s gravity is natural and cannot be generated by satellites. Artificial satellites serve roles in communication, navigation, research, and defense.
338. Which of the following best describes natural satellites?
ⓐ. Man-made objects launched into orbit
ⓑ. Celestial bodies that naturally orbit planets
ⓒ. Only bodies that emit light
ⓓ. Spacecraft sent to study outer planets
Correct Answer: Celestial bodies that naturally orbit planets
Explanation: Natural satellites are naturally occurring moons or small celestial bodies that orbit planets due to gravity.
339. Which of the following is an example of a navigation satellite?
ⓐ. GPS
ⓑ. Voyager 1
ⓒ. Hubble Telescope
ⓓ. Cartosat
Correct Answer: GPS
Explanation: GPS (Global Positioning System) satellites form a network of navigation satellites used for determining position and time information worldwide.
340. What was India’s first artificial satellite?
ⓐ. INSAT-1
ⓑ. Cartosat-1
ⓒ. Aryabhata
ⓓ. GSAT-1
Correct Answer: Aryabhata
Explanation: Aryabhata was India’s first artificial satellite, launched in 1975 by the Soviet Union. It was mainly designed for scientific experiments in space.
341. Which force is responsible for keeping a satellite in its orbit around Earth?
ⓐ. Magnetic force
ⓑ. Centrifugal force
ⓒ. Gravitational force
ⓓ. Nuclear force
Correct Answer: Gravitational force
Explanation: Gravitational force acts as the centripetal force that keeps satellites in orbit around Earth. Without it, satellites would move in a straight line due to inertia.
342. Which condition must be satisfied for a satellite to remain in circular orbit?
ⓐ. \(\frac{GMm}{r^2} = \frac{mv^2}{r}\)
ⓑ. \(GM = R^2\)
ⓒ. \(v = \sqrt{\frac{GM}{R^2}}\)
ⓓ. \(\frac{mv}{r} = 0\)
Correct Answer: \(\frac{GMm}{r^2} = \frac{mv^2}{r}\)
Explanation: For circular orbit, gravitational force \(\frac{GMm}{r^2}\) must equal centripetal force \(\frac{mv^2}{r}\).
343. The orbital velocity of a satellite at a distance \(r\) from Earth’s center is:
ⓐ. \(v = \sqrt{\frac{GM}{r}}\)
ⓑ. \(v = \sqrt{\frac{2GM}{r}}\)
ⓒ. \(v = \frac{GM}{r^2}\)
ⓓ. \(v = \sqrt{gr}\)
Correct Answer: \(v = \sqrt{\frac{GM}{r}}\)
Explanation: From \(\frac{GMm}{r^2} = \frac{mv^2}{r}\), solving gives \(v = \sqrt{\tfrac{GM}{r}}\).
344. If the orbital radius of a satellite increases, its orbital velocity will:
ⓐ. Increase
ⓑ. Decrease
ⓒ. Remain constant
ⓓ. First increase, then decrease
Correct Answer: Decrease
Explanation: Since \(v = \sqrt{\tfrac{GM}{r}}\), as \(r\) increases, orbital velocity decreases.
345. Which energy is constant for a satellite in circular orbit?
ⓐ. Kinetic energy only
ⓑ. Potential energy only
ⓒ. Total mechanical energy
ⓓ. Both kinetic and potential energies
Correct Answer: Total mechanical energy
Explanation: In circular orbit, kinetic and potential energies vary with radius, but their sum (total energy) remains constant.
346. The time period of a satellite orbiting Earth is given by:
ⓐ. \(T = 2\pi \sqrt{\frac{r}{g}}\)
ⓑ. \(T = 2\pi \sqrt{\frac{r^3}{GM}}\)
ⓒ. \(T = \sqrt{\frac{GM}{r}}\)
ⓓ. \(T = \frac{2\pi r}{GM}\)
Correct Answer: \(T = 2\pi \sqrt{\frac{r^3}{GM}}\)
Explanation: Using \(v = \sqrt{\tfrac{GM}{r}}\) and \(T = \tfrac{2\pi r}{v}\), we get \(T = 2\pi \sqrt{\tfrac{r^3}{GM}}\), consistent with Kepler’s Third Law.
347. A satellite is launched into a higher orbit. Which of the following increases?
ⓐ. Orbital velocity
ⓑ. Gravitational force on satellite
ⓒ. Orbital period
ⓓ. Acceleration due to gravity
Correct Answer: Orbital period
Explanation: At higher orbits, velocity decreases and period increases because \(T = 2\pi \sqrt{\tfrac{r^3}{GM}}\).
348. What is the relation between orbital speed and escape speed at the same distance \(r\)?
ⓐ. \(v_o = v_e\)
ⓑ. \(v_o = \sqrt{2} v_e\)
ⓒ. \(v_o = \frac{v_e}{\sqrt{2}}\)
ⓓ. \(v_o = 2 v_e\)
Correct Answer: \(v_o = \frac{v_e}{\sqrt{2}}\)
Explanation: Escape speed is \(v_e = \sqrt{\tfrac{2GM}{r}}\) while orbital speed is \(v_o = \sqrt{\tfrac{GM}{r}}\). Thus, \(v_o = \tfrac{v_e}{\sqrt{2}}\).
349. Why don’t satellites fall back to Earth?
ⓐ. Because there is no gravity in space
ⓑ. Because they are beyond Earth’s gravitational field
ⓒ. Because their tangential velocity balances gravitational pull
ⓓ. Because they rotate faster than Earth
Correct Answer: Because their tangential velocity balances gravitational pull
Explanation: Satellites are in free fall, but their horizontal velocity ensures they keep missing Earth, resulting in continuous orbit.
350. What happens to the centripetal acceleration of a satellite as its orbital radius increases?
ⓐ. It increases
ⓑ. It decreases
ⓒ. It remains constant
ⓓ. It becomes infinite
Correct Answer: It decreases
Explanation: Centripetal acceleration is \(a = \frac{v^2}{r} = \frac{GM}{r^2}\). As \(r\) increases, acceleration decreases.
351. What does the altitude of a satellite’s orbit represent?
ⓐ. The distance from Earth’s surface to the satellite
ⓑ. The distance from Earth’s center to the satellite
ⓒ. The velocity of the satellite in orbit
ⓓ. The angle between orbital plane and equator
Correct Answer: The distance from Earth’s surface to the satellite
Explanation: Altitude is measured from Earth’s surface, not the center. For orbital equations, however, we often use the radius \(r = R_E + h\), where \(R_E\) is Earth’s radius and \(h\) is altitude.
352. If altitude of a satellite increases, its orbital period:
ⓐ. Decreases
ⓑ. Increases
ⓒ. Remains constant
ⓓ. Becomes zero
Correct Answer: Increases
Explanation: The orbital period is \(T = 2\pi \sqrt{\tfrac{r^3}{GM}}\). Larger \(r\) (higher altitude) increases \(T\).
353. What is the orbital inclination of a satellite?
ⓐ. Angle between orbital velocity and Earth’s radius
ⓑ. Angle between the orbital plane and Earth’s equatorial plane
ⓒ. Angle between the satellite and Earth’s axis
ⓓ. Angle between gravitational force and centrifugal force
Correct Answer: Angle between the orbital plane and Earth’s equatorial plane
Explanation: Orbital inclination is measured relative to Earth’s equator. It defines whether an orbit is equatorial, polar, or inclined.
354. A satellite with an inclination of \(0^\circ\) is in:
ⓐ. Polar orbit
ⓑ. Equatorial orbit
ⓒ. Geosynchronous orbit
ⓓ. Sun-synchronous orbit
Correct Answer: Equatorial orbit
Explanation: At \(0^\circ\) inclination, the orbit lies exactly in Earth’s equatorial plane. Polar orbits have \(90^\circ\) inclination.
355. A polar orbit has an inclination of approximately:
ⓐ. \(0^\circ\)
ⓑ. \(23.5^\circ\)
ⓒ. \(60^\circ\)
ⓓ. \(90^\circ\)
Correct Answer: \(90^\circ\)
Explanation: Polar orbits pass over Earth’s poles, meaning their orbital plane is perpendicular to the equatorial plane.
356. If a satellite’s orbital radius is doubled, its orbital period increases by a factor of:
ⓐ. 2
ⓑ. \(\sqrt{2}\)
ⓒ. \(2^{3/2}\)
ⓓ. 4
Correct Answer: \(2^{3/2}\)
Explanation: From \(T \propto r^{3/2}\), doubling \(r\) multiplies \(T\) by \(2^{3/2} \approx 2.83\).
357. Which of the following determines the speed of a satellite in orbit?
ⓐ. Mass of the satellite only
ⓑ. Radius of orbit and Earth’s mass
ⓒ. Altitude only
ⓓ. Inclination only
Correct Answer: Radius of orbit and Earth’s mass
Explanation: Orbital speed is given by \(v = \sqrt{\tfrac{GM}{r}}\). It depends on Earth’s mass and orbital radius, not satellite mass.
358. The time period of a satellite depends on:
ⓐ. Its mass and altitude
ⓑ. Earth’s mass and orbital radius
ⓒ. Earth’s radius only
ⓓ. Inclination only
Correct Answer: Earth’s mass and orbital radius
Explanation: From \(T = 2\pi \sqrt{\tfrac{r^3}{GM}}\), the time period is determined by Earth’s mass \(M\) and orbital radius \(r\), independent of the satellite’s mass.
359. What is the altitude of a geostationary satellite approximately?
ⓐ. 3600 km
ⓑ. 8600 km
ⓒ. 22,300 km
ⓓ. 42,200 km
Correct Answer: 22,300 km
Explanation: Geostationary satellites orbit at an altitude of about 35,786 km above Earth’s surface, i.e. \~42,200 km from Earth’s center.
360. If the inclination of a satellite is not zero, the satellite will:
ⓐ. Remain fixed above one point on Earth
ⓑ. Drift north and south of the equator during orbit
ⓒ. Not complete one orbit
ⓓ. Fall back to Earth
Correct Answer: Drift north and south of the equator during orbit
Explanation: Only geostationary satellites (inclination = 0°) remain fixed relative to Earth. Inclined orbits cause the satellite to oscillate across latitudes as Earth rotates beneath.
361. Which of the following is the primary use of communication satellites?
ⓐ. Measuring Earth’s gravity
ⓑ. Transmitting television, internet, and telephone signals
ⓒ. Observing black holes
ⓓ. Launching rockets into space
Correct Answer: Transmitting television, internet, and telephone signals
Explanation: Communication satellites relay signals between ground stations, allowing long-distance communication. They use transponders to receive, amplify, and retransmit signals.
362. How do weather satellites help in forecasting?
ⓐ. By monitoring changes in Earth’s magnetic field
ⓑ. By capturing cloud patterns, temperature, and wind movements
ⓒ. By measuring earthquake activity
ⓓ. By reflecting sunlight back to Earth
Correct Answer: By capturing cloud patterns, temperature, and wind movements
Explanation: Weather satellites provide real-time images and data on atmospheric conditions, enabling predictions of storms, rainfall, cyclones, and climate patterns.
363. Which type of satellite is used for navigation purposes?
ⓐ. INSAT
ⓑ. GPS
ⓒ. Hubble
ⓓ. Cartosat
Correct Answer: GPS
Explanation: GPS (Global Positioning System) satellites provide precise location and timing services, widely used in smartphones, vehicles, and aviation.
364. Which Indian satellite series is primarily used for communication?
ⓐ. INSAT
ⓑ. IRS
ⓒ. Cartosat
ⓓ. Hubble
Correct Answer: INSAT
Explanation: INSAT (Indian National Satellite System) is designed for telecommunication, broadcasting, weather monitoring, and search-and-rescue operations.
365. Which satellite helps in Earth resource management such as agriculture and forestry?
ⓐ. IRS (Indian Remote Sensing Satellite)
ⓑ. INSAT
ⓒ. Hubble
ⓓ. Voyager 2
Correct Answer: IRS (Indian Remote Sensing Satellite)
Explanation: IRS satellites provide imagery and data useful in agriculture, forestry, water resources, and environmental monitoring.
366. Which of the following satellites is used to study the universe beyond Earth’s atmosphere?
ⓐ. Aryabhata
ⓑ. GSAT
ⓒ. Hubble Space Telescope
ⓓ. INSAT
Correct Answer: Hubble Space Telescope
Explanation: Hubble is an astronomical observatory satellite that studies stars, galaxies, and planets beyond Earth’s atmosphere with high-resolution imaging.
367. Why are geostationary satellites ideal for communication?
ⓐ. They move faster than Earth’s rotation
ⓑ. They remain fixed over one point on Earth
ⓒ. They do not require power
ⓓ. They orbit near the poles
Correct Answer: They remain fixed over one point on Earth
Explanation: Geostationary satellites orbit Earth once every 24 hours, appearing stationary above a location, making them ideal for continuous communication links.
368. Which satellite application is most useful in predicting cyclones and hurricanes?
ⓐ. Communication satellites
ⓑ. Navigation satellites
ⓒ. Weather satellites
ⓓ. Astronomical satellites
Correct Answer: Weather satellites
Explanation: Weather satellites monitor cloud structures, ocean temperatures, and wind speeds, helping meteorologists predict cyclones and hurricanes in advance.
369. What is the primary function of navigation satellites in aviation?
ⓐ. To measure air pressure in cockpits
ⓑ. To provide accurate position and route information
ⓒ. To guide aircraft engines
ⓓ. To track black holes
Correct Answer: To provide accurate position and route information
Explanation: Navigation satellites like GPS, GLONASS, and Galileo allow aircraft to navigate safely by giving precise real-time location and route tracking.
370. Which of the following is a combined application of communication and weather satellites?
ⓐ. Telemedicine in remote areas
ⓑ. Measuring gravitational constant
ⓒ. Studying nuclear reactions in stars
ⓓ. Launching space probes
Correct Answer: Telemedicine in remote areas
Explanation: Communication satellites enable real-time consultations between doctors and patients in remote areas, while weather satellites provide climate data for planning medical support in disaster-affected regions.
371. What is the expression for the kinetic energy of a satellite in circular orbit of radius \(r\)?
ⓐ. \(KE = \frac{GMm}{2r}\)
ⓑ. \(KE = \frac{GMm}{r}\)
ⓒ. \(KE = \frac{GM}{2r}\)
ⓓ. \(KE = \frac{Gm^2}{2r}\)
Correct Answer: \(KE = \frac{GMm}{2r}\)
Explanation: Orbital velocity is \(v = \sqrt{\tfrac{GM}{r}}\). Kinetic energy is \(\tfrac{1}{2}mv^2 = \tfrac{1}{2}m \cdot \tfrac{GM}{r} = \tfrac{GMm}{2r}\).
372. If the orbital radius of a satellite doubles, its kinetic energy becomes:
ⓐ. Twice as large
ⓑ. Half as large
ⓒ. One-fourth as large
ⓓ. Remains unchanged
Correct Answer: Half as large
Explanation: \(KE \propto \tfrac{1}{r}\). Doubling \(r\) reduces \(KE\) to half its previous value.
373. Which of the following is true about kinetic energy of an orbiting satellite?
ⓐ. It is equal to its gravitational potential energy
ⓑ. It is always positive
ⓒ. It depends on the mass of the planet only
ⓓ. It increases with altitude
Correct Answer: It is always positive
Explanation: Kinetic energy is always positive, while gravitational potential energy is negative. For satellites, \(KE = -\tfrac{1}{2} U\), ensuring total energy is negative (bound system).
374. How is kinetic energy of a satellite related to its total energy?
ⓐ. \(KE = E\)
ⓑ. \(KE = -E\)
ⓒ. \(KE = 2E\)
ⓓ. \(KE = \frac{E}{2}\)
Correct Answer: \(KE = -E\)
Explanation: For a satellite, total energy \(E = KE + U = -\tfrac{GMm}{2r}\). Since \(KE = \tfrac{GMm}{2r}\), we find \(KE = -E\).
375. The kinetic energy of a satellite depends on:
ⓐ. Only the satellite’s mass
ⓑ. Only Earth’s mass
ⓒ. Both satellite mass and orbital radius
ⓓ. Altitude only
Correct Answer: Both satellite mass and orbital radius
Explanation: \(KE = \tfrac{GMm}{2r}\), which depends on gravitational constant \(G\), Earth’s mass \(M\), satellite’s mass \(m\), and orbital radius \(r\).
376. For two satellites of different masses in the same orbit, their kinetic energies are:
ⓐ. Equal
ⓑ. Proportional to their masses
ⓒ. Independent of their masses
ⓓ. Proportional to the square of their masses
Correct Answer: Proportional to their masses
Explanation: Since \(KE = \tfrac{GMm}{2r}\), it is directly proportional to the satellite’s mass \(m\). Thus, a heavier satellite in the same orbit has higher kinetic energy.
377. What happens to the kinetic energy of a satellite when it is moved to a higher orbit?
ⓐ. It increases
ⓑ. It decreases
ⓒ. It remains constant
ⓓ. It becomes zero
Correct Answer: It decreases
Explanation: In higher orbit, orbital speed is lower, so kinetic energy decreases according to \(KE = \tfrac{GMm}{2r}\).
378. If a satellite of mass \(m\) is orbiting Earth of mass \(M\), its kinetic energy can be expressed as:
ⓐ. \(\frac{GM^2}{2r}\)
ⓑ. \(\frac{GMm}{2r}\)
ⓒ. \(\frac{Gm^2}{2r}\)
ⓓ. \(\frac{M}{2Gm}\)
Correct Answer: \(\frac{GMm}{2r}\)
Explanation: Using orbital mechanics, kinetic energy of a satellite is \(KE = \frac{GMm}{2r}\), depending on both masses and orbital radius.
379. The ratio of kinetic energy to gravitational potential energy of a satellite in orbit is:
ⓐ. \(1\)
ⓑ. \(\frac{1}{2}\)
ⓒ. \(-\frac{1}{2}\)
ⓓ. \(-1\)
Correct Answer: \(-\frac{1}{2}\)
Explanation: Potential energy is \(U = -\frac{GMm}{r}\), kinetic energy is \(KE = \frac{GMm}{2r}\). Thus, \(\frac{KE}{U} = -\tfrac{1}{2}\).
380. Which principle explains why kinetic energy decreases with increasing altitude of satellites?
ⓐ. Newton’s First Law
ⓑ. Conservation of Energy
ⓒ. Conservation of Angular Momentum
ⓓ. Law of Gravitation
Correct Answer: Conservation of Energy
Explanation: As satellites are moved to higher orbits, work must be done against gravity, which reduces their orbital velocity and hence their kinetic energy while maintaining total energy conservation.
381. What is the expression for the potential energy of a satellite of mass \(m\) orbiting Earth at radius \(r\)?
ⓐ. \(U = \frac{GMm}{r}\)
ⓑ. \(U = -\frac{GMm}{r}\)
ⓒ. \(U = -\frac{GM}{r^2}\)
ⓓ. \(U = \frac{GMm}{2r}\)
Correct Answer: \(U = -\frac{GMm}{r}\)
Explanation: Gravitational potential energy is negative because gravity is an attractive force. The expression \(U = -\frac{GMm}{r}\) represents the work required to bring the satellite from infinity to radius \(r\).
382. Why is the gravitational potential energy of an orbiting satellite negative?
ⓐ. Because energy is lost due to friction
ⓑ. Because infinity is taken as reference with zero potential
ⓒ. Because kinetic energy is greater than potential energy
ⓓ. Because satellites emit energy in orbit
Correct Answer: Because infinity is taken as reference with zero potential
Explanation: By convention, potential energy at infinity is zero. Since work must be done against gravity to move a body from radius \(r\) to infinity, potential energy at finite distance is negative.
383. How does potential energy of a satellite vary with orbital radius \(r\)?
ⓐ. \(U \propto r\)
ⓑ. \(U \propto \frac{1}{r^2}\)
ⓒ. \(U \propto -\frac{1}{r}\)
ⓓ. \(U \propto \sqrt{r}\)
Correct Answer: \(U \propto -\frac{1}{r}\)
Explanation: The formula \(U = -\frac{GMm}{r}\) shows that potential energy decreases in magnitude as \(r\) decreases, but becomes less negative at larger distances.
384. Which of the following statements is true about the potential energy of a satellite?
ⓐ. It increases (less negative) as altitude increases
ⓑ. It is always positive in orbit
ⓒ. It is independent of satellite mass
ⓓ. It remains constant regardless of radius
Correct Answer: It increases (less negative) as altitude increases
Explanation: Since \(U = -\frac{GMm}{r}\), larger \(r\) means potential energy moves toward zero (less negative).
385. If a satellite moves to twice its orbital radius, its potential energy becomes:
ⓐ. Twice as large (less negative)
ⓑ. Half as large (less negative)
ⓒ. Double in magnitude (more negative)
ⓓ. Zero
Correct Answer: Half as large (less negative)
Explanation: Potential energy is inversely proportional to \(r\). Doubling \(r\) reduces the magnitude of \(U\) by half, making it less negative.
386. The potential energy of a satellite is related to its kinetic energy as:
ⓐ. \(U = KE\)
ⓑ. \(U = -2KE\)
ⓒ. \(U = -KE\)
ⓓ. \(U = 2KE\)
Correct Answer: \(U = -2KE\)
Explanation: For circular orbit, \(KE = \frac{GMm}{2r}\) and \(U = -\frac{GMm}{r}\). Clearly, \(U = -2KE\).
387. What is the total mechanical energy \(E\) of a satellite in terms of its potential energy \(U\)?
ⓐ. \(E = U\)
ⓑ. \(E = \frac{U}{2}\)
ⓒ. \(E = 2U\)
ⓓ. \(E = -U\)
Correct Answer: \(E = \frac{U}{2}\)
Explanation: Since \(E = KE + U\) and \(U = -2KE\), then \(E = KE + U = KE – 2KE = -KE = \frac{U}{2}\).
388. If a satellite of mass \(m\) is placed at infinity, its potential energy will be:
ⓐ. Zero
ⓑ. Positive
ⓒ. Negative infinity
ⓓ. Equal to kinetic energy
Correct Answer: Zero
Explanation: By definition, gravitational potential energy at infinity is taken as zero reference point.
389. Which factor directly affects the magnitude of potential energy of an orbiting satellite?
ⓐ. Only Earth’s mass
ⓑ. Only satellite’s velocity
ⓒ. Both satellite mass and orbital radius
ⓓ. Inclination of orbit
Correct Answer: Both satellite mass and orbital radius
Explanation: \(U = -\frac{GMm}{r}\), so it depends on both Earth’s mass \(M\), satellite’s mass \(m\), and orbital radius \(r\).
390. If the gravitational constant \(G\) were larger, how would it affect the potential energy of a satellite?
ⓐ. No effect
ⓑ. Potential energy would become more negative
ⓒ. Potential energy would become positive
ⓓ. Potential energy would become zero
Correct Answer: Potential energy would become more negative
Explanation: Since \(U = -\frac{GMm}{r}\), increasing \(G\) increases the magnitude of gravitational attraction, thus making potential energy more negative.
391. What is the expression for the total mechanical energy of a satellite in circular orbit of radius \(r\)?
ⓐ. \(E = \frac{GMm}{r}\)
ⓑ. \(E = -\frac{GMm}{2r}\)
ⓒ. \(E = -\frac{GMm}{r}\)
ⓓ. \(E = \frac{GMm}{2r}\)
Correct Answer: \(E = -\frac{GMm}{2r}\)
Explanation: Total mechanical energy is the sum of kinetic and potential energies. Since \(KE = \frac{GMm}{2r}\) and \(U = -\frac{GMm}{r}\), we get \(E = KE + U = -\frac{GMm}{2r}\).
392. Why is the total mechanical energy of a satellite always negative?
ⓐ. Because kinetic energy is zero
ⓑ. Because potential energy dominates and is negative
ⓒ. Because satellites radiate energy
ⓓ. Because energy is measured relative to Sun
Correct Answer: Because potential energy dominates and is negative
Explanation: Gravitational potential energy is twice the kinetic energy in magnitude and negative. Therefore, the sum remains negative, indicating a bound system.
393. If the total mechanical energy of a satellite becomes zero, what will happen?
ⓐ. The satellite remains in circular orbit
ⓑ. The satellite escapes Earth’s gravitational field
ⓒ. The satellite’s speed becomes infinite
ⓓ. The satellite falls into Earth
Correct Answer: The satellite escapes Earth’s gravitational field
Explanation: \(E = 0\) corresponds to escape velocity, where the satellite has just enough energy to move infinitely far from Earth.
394. How is total energy of a satellite related to its kinetic energy?
ⓐ. \(E = KE\)
ⓑ. \(E = -KE\)
ⓒ. \(E = 2KE\)
ⓓ. \(E = -2KE\)
Correct Answer: \(E = -KE\)
Explanation: For circular orbit, \(KE = \frac{GMm}{2r}\) and \(E = -\frac{GMm}{2r}\). Hence, \(E = -KE\).
395. How is total energy of a satellite related to its potential energy?
ⓐ. \(E = U\)
ⓑ. \(E = -U\)
ⓒ. \(E = \frac{U}{2}\)
ⓓ. \(E = 2U\)
Correct Answer: \(E = \frac{U}{2}\)
Explanation: Since \(U = -\frac{GMm}{r}\) and \(E = -\frac{GMm}{2r}\), we find \(E = \frac{U}{2}\).
396. If orbital radius of a satellite is increased, the total mechanical energy becomes:
ⓐ. More negative
ⓑ. Less negative
ⓒ. Zero
ⓓ. Positive
Correct Answer: Less negative
Explanation: \(E = -\frac{GMm}{2r}\). Increasing \(r\) reduces magnitude of \(E\), making it less negative, meaning the satellite is less tightly bound.
397. Which of the following correctly represents the distribution of energy for a satellite in circular orbit?
ⓐ. \(KE = U\)
ⓑ. \(KE = -U\)
ⓒ. \(KE = -E\), \(U = 2E\)
ⓓ. \(KE = E\), \(U = -E\)
Correct Answer: \(KE = -E\), \(U = 2E\)
Explanation: From orbital energy relations: \(KE = -E\), \(U = 2E\). This ensures \(E = KE + U = -KE\).
398. For an orbiting satellite, what does the negative sign of total energy indicate?
ⓐ. Energy is lost to space
ⓑ. The satellite is bound to Earth
ⓒ. The satellite has infinite velocity
ⓓ. The satellite cannot exist in orbit
Correct Answer: The satellite is bound to Earth
Explanation: A negative total energy implies the satellite is gravitationally bound and cannot escape unless additional energy is provided.
399. If the total mechanical energy of a satellite is doubled (less negative), what happens to its orbit?
ⓐ. Orbit radius decreases
ⓑ. Orbit radius increases
ⓒ. Orbit becomes unstable and satellite falls
ⓓ. Orbit remains unchanged
Correct Answer: Orbit radius increases
Explanation: Since \(E = -\frac{GMm}{2r}\), if \(E\) becomes less negative, \(r\) must increase. Thus, the satellite moves to a higher orbit.
400. The total mechanical energy per unit mass of a satellite in circular orbit is:
ⓐ. \(-\frac{GM}{r}\)
ⓑ. \(-\frac{GM}{2r}\)
ⓒ. \(\frac{GM}{2r}\)
ⓓ. \(\frac{GM}{r}\)
Correct Answer: \(-\frac{GM}{2r}\)
Explanation: Dividing total energy \(E = -\frac{GMm}{2r}\) by mass \(m\), we obtain energy per unit mass \(\epsilon = -\frac{GM}{2r}\). This is useful for analyzing satellite dynamics without dependence on satellite mass.
The chapter Gravitation is an integral part of NCERT/CBSE Class 11 Physics syllabus, forming the base for understanding planetary motion and space science. It is equally important for scoring in board exams as well as in competitive exams like JEE, NEET, and other entrance tests. This complete set includes 580 MCQs with answers and explanations, divided into 6 parts for easy practice. In this fourth section, you will get access to another 100 MCQs with solutions, perfect for building accuracy and speed before exams.
- 👉 Total MCQs in this chapter: 580.
- 👉 This page contains: Fourth set of 100 solved MCQs.
- 👉 Great for board exam preparation and competitive practice.
- 👉 Browse more chapters or subjects via the top navigation bar.
- 👉 Next set of questions is available in Part 5 above.