301. Tidal effects are mainly produced because the gravitational force of the Moon on Earth
ⓐ. varies slightly across Earth’s diameter
ⓑ. is identical at every point on Earth
ⓒ. acts only on oceans and not on land
ⓓ. becomes repulsive on Earth’s far side
Correct Answer: varies slightly across Earth’s diameter
Explanation: Tides are related to the variation of gravitational pull across Earth’s size. The side of Earth nearer the Moon experiences a slightly stronger lunar gravitational pull than Earth’s centre, while the far side experiences a slightly weaker pull. This difference in pull is called a gravitational gradient effect. The Moon’s gravity acts on all parts of Earth, not only on ocean water. Oceans show the effect more visibly because water can move more easily than solid land.
302. The term gravitational gradient in the context of tides refers to
ⓐ. Earth’s speed changes in its orbit around the Sun
ⓑ. field or force changes with position
ⓒ. Earth’s density changes from place to place
ⓓ. the constant \(G\) changes from place to place
Correct Answer: field or force changes with position
Explanation: A gravitational gradient means that the gravitational field is not exactly the same at nearby points. In tides, the Moon’s gravitational field differs slightly between Earth’s near side, centre, and far side. This small difference is more important for tides than the average pull on Earth as a whole. A uniform gravitational field would accelerate all parts equally and would not stretch Earth in the same way. The gradient idea depends on variation with position, not on the numerical value of \(G\) alone.
303. The Moon produces noticeable tides on Earth mainly because
ⓐ. the Moon has greater mass than the Sun and pulls uniformly
ⓑ. the Moon changes \(G\) across Earth because it is close
ⓒ. the Moon is close, so its field varies across Earth
ⓓ. lunar gravity repels seawater on the far side of Earth
Correct Answer: the Moon is close, so its field varies across Earth
Explanation: The Sun has much greater mass than the Moon and exerts a strong gravitational pull on Earth. However, tidal effects depend strongly on how much the gravitational pull changes across Earth’s diameter. The Moon is much closer to Earth, so its field has a relatively larger difference between the near side and far side of Earth. This makes the Moon a major cause of tides. The value of \(G\) does not change between day and night.
304. A common statement says, “Tides happen because the Moon pulls only the water nearest to it.” The best correction is that the Moon
ⓐ. pulls all parts of Earth by different amounts
ⓑ. pulls only the nearest ocean and not the rest of Earth
ⓒ. has no effect on solid Earth
ⓓ. repels the far-side ocean instead of attracting it
Correct Answer: pulls all parts of Earth by different amounts
Explanation: The Moon’s gravitational attraction acts on the whole Earth. The key point is that the attraction is not exactly equal at all parts of Earth. The near side is pulled slightly more strongly than Earth’s centre, while the far side is pulled slightly less strongly. This difference produces the stretching tendency associated with tides. Describing tides as a pull on only one patch of water misses the whole-Earth nature of the gravitational interaction.
305. Use the arrangement described below. The Moon is to the right of Earth. Point P is on Earth’s surface nearest the Moon, Point Q is at Earth’s centre, and Point R is on the far side. The Moon’s gravitational pull is strongest at
ⓐ. Point P
ⓑ. all three points equally
ⓒ. Point R
ⓓ. Point Q
Correct Answer: Point P
Explanation: Gravitational force due to the Moon decreases with distance from the Moon. Point P is nearest to the Moon, so the Moon’s pull is strongest there. Point Q is farther from the Moon than P, so the pull at Q is slightly weaker. Point R is farthest from the Moon, so the pull at R is weakest. The tidal effect comes from these small differences, not from a complete absence of gravity anywhere on Earth.
306. The following table compares three points on Earth along the Earth-Moon line.
| Point | Position relative to Moon | Lunar gravitational pull |
|---|
| P | Near side of Earth | Strongest among the three |
| Q | Earth’s centre | Intermediate |
| R | Far side of Earth | Weakest among the three |
The table supports the idea that tides are related to
ⓐ. changing universal constant across Earth’s diameter
ⓑ. same lunar pull on near and far sides of Earth
ⓒ. different lunar pulls at different Earth points
ⓓ. absence of gravity at Earth’s near-side surface
Correct Answer: different lunar pulls at different Earth points
Explanation: The table shows that the Moon’s pull is not exactly the same at all three points. This difference occurs because the points are at slightly different distances from the Moon. A tidal effect is linked with this variation of pull across Earth, not just with the average attraction between Earth and the Moon. Earth’s centre does not have zero lunar attraction in this situation. The constant \(G\) remains the same; the distances are what change the pull.
307. Assertion: Tides are more closely connected with the difference in gravitational pull across Earth than with the same pull applied equally to all parts of Earth. Reason: Equal gravitational acceleration of every part would not produce stretching of Earth’s oceans relative to Earth.
ⓐ. The assertion is true, but the reason is false
ⓑ. Both assertion and reason are true, and the reason explains the assertion
ⓒ. Both assertion and reason are true, but the reason does not explain the assertion
ⓓ. The assertion is false, but the reason is true
Correct Answer: Both assertion and reason are true, and the reason explains the assertion
Explanation: The assertion is true because tidal behaviour depends on nonuniform gravitational pull. If every part of Earth experienced exactly the same acceleration due to the Moon, Earth and its oceans would simply accelerate together without the same relative stretching effect. The reason correctly explains why the difference across Earth matters. The near side, centre, and far side do not receive exactly identical lunar pulls. Tides are therefore a gradient effect rather than merely a result of the Moon pulling Earth as a whole.
308. A compact passage says: “The Sun pulls Earth more strongly than the Moon does, but the Moon has a larger tidal influence than one might expect from its mass alone.” The most suitable explanation is that tidal influence depends strongly on
ⓐ. whether the attracting body emits its own light
ⓑ. brightness variation across Earth, larger for closer bodies
ⓒ. mass of seawater alone instead of distance variation
ⓓ. pull variation across Earth, larger for closer bodies
Correct Answer: pull variation across Earth, larger for closer bodies
Explanation: The Sun’s total pull on Earth is large because the Sun has enormous mass. Tidal influence, however, depends on how much the gravitational pull changes from one side of Earth to the other. A closer body produces a larger fractional change in distance across Earth’s diameter. The Moon is much closer than the Sun, so its gravitational gradient across Earth is very important for tides. Light emission has no role in Newtonian gravitational attraction.
309. For two fixed masses, a graph of gravitational force \(F\) against \(\frac{1}{r^2}\) should be
ⓐ. a straight line not through the origin
ⓑ. a straight line through the origin
ⓒ. a curve decreasing with \(r^2\)
ⓓ. a horizontal line above the axis
Correct Answer: a straight line through the origin
Explanation: Newton’s law gives \(F=G\frac{m_1m_2}{r^2}\). If \(m_1\), \(m_2\), and \(G\) are constant, then \(F\propto\frac{1}{r^2}\). A graph of \(F\) against \(\frac{1}{r^2}\) therefore has the form \(y=kx\). Such a graph is a straight line through the origin. This differs from a graph of \(F\) against \(r\), which is an inverse-square curve.
310. In a graph of \(F\) against \(\frac{1}{r^2}\) for two fixed masses \(m_1\) and \(m_2\), the slope is
ⓐ. \(\frac{m_1m_2}{G}\)
ⓑ. \(G\frac{m_1}{m_2}\)
ⓒ. \(\frac{G}{m_1m_2}\)
ⓓ. \(Gm_1m_2\)
Correct Answer: \(Gm_1m_2\)
Explanation: \( \textbf{Start with Newton’s law:} \)
\[
F=G\frac{m_1m_2}{r^2}
\]
Let the horizontal-axis variable be
\[
x=\frac{1}{r^2}
\]
Then the force becomes
\[
F=Gm_1m_2x
\]
This has the straight-line form
\[
y=kx
\]
where
\[
k=Gm_1m_2
\]
So the slope is
\[
Gm_1m_2
\]
\( \textbf{Final answer:} \) \(Gm_1m_2\), because the masses and \(G\) multiply the plotted variable \(\frac{1}{r^2}\).
311. The graph of gravitational field \(g(r)\) for a uniform solid Earth is divided into two regions: Region P has \(0\le r\le R_E\), and Region Q has \(r\ge R_E\). The correct description is
ⓐ. P is zero everywhere, Q is zero everywhere
ⓑ. both P and Q are horizontal
ⓒ. P is linear increasing, Q is inverse-square decreasing
ⓓ. P is inverse-square decreasing, Q is linear increasing
Correct Answer: P is linear increasing, Q is inverse-square decreasing
Explanation: Inside a uniform solid Earth, only the mass enclosed within radius \(r\) effectively contributes to the field at that point. This gives \(g(r)\propto r\) for \(0\le r\le R_E\), so the graph rises linearly from the centre to the surface. Outside Earth, the whole mass acts as if concentrated at the centre, so \(g(r)=\frac{GM_E}{r^2}\). Therefore the outside graph decreases as an inverse-square curve. The surface is the joining point of the two behaviours.
312. The slope of the \(g(r)\) graph inside a uniform Earth, when \(g(r)=g_s\frac{r}{R_E}\), is
ⓐ. \(g_sR_E\)
ⓑ. \(\frac{g_s}{R_E}\)
ⓒ. \(\frac{R_E}{g_s}\)
ⓓ. \(g_sR_E^2\)
Correct Answer: \(\frac{g_s}{R_E}\)
Explanation: \( \textbf{Interior relation:} \)
\[
g(r)=g_s\frac{r}{R_E}
\]
Rewrite it as
\[
g(r)=\left(\frac{g_s}{R_E}\right)r
\]
This is in the straight-line form
\[
y=mx
\]
where \(y=g(r)\), \(x=r\), and the slope is
\[
m=\frac{g_s}{R_E}
\]
\( \textbf{Final answer:} \) \(\frac{g_s}{R_E}\), with unit \(\text{s}^{-2}\) because it is acceleration divided by length.
313. A graph of gravitational potential \(V\) against distance \(r\) from an isolated mass is negative and approaches \(0\) from below. A graph of gravitational field magnitude \(g\) against \(r\) is positive and decreases. The main reason for this difference is that
ⓐ. potential is scalar and negative; field magnitude is positive
ⓑ. potential is vector with direction, while field has only magnitude
ⓒ. field becomes negative because the source mass is negative
ⓓ. potential is independent of distance from the source mass
Correct Answer: potential is scalar and negative; field magnitude is positive
Explanation: For a point mass, gravitational potential is \(V=-\frac{GM}{r}\). It is negative because the zero reference is chosen at infinity and gravity is attractive. The gravitational field magnitude is \(g=\frac{GM}{r^2}\), which is written as a positive magnitude. Field as a vector has direction inward, but its magnitude is positive. The two graphs differ in sign and power of \(r\), so they should not be treated as the same curve.
314. A graph of \(V\) against \(r\) for a point mass becomes flatter as \(r\) increases. This means the gravitational field magnitude
ⓐ. remains constant at all distances
ⓑ. increases with distance
ⓒ. becomes negative in magnitude
ⓓ. decreases with distance
Correct Answer: decreases with distance
Explanation: The field is related to the radial slope of potential by \(g_r=-\frac{dV}{dr}\). If the \(V-r\) graph becomes flatter, the magnitude of its slope decreases. A smaller slope magnitude corresponds to a smaller gravitational field magnitude. This agrees with \(g=\frac{GM}{r^2}\), which decreases as \(r\) increases. The field direction remains inward, but the question asks about magnitude.
315. The following table lists graph relations for gravitation.
| Graph | Expected relation |
|---|
| P. \(F\) versus \(\frac{1}{r^2}\) | Straight line through origin |
| Q. \(T^2\) versus \(r^3\) | Straight line through origin |
| R. \(V\) versus \(\frac{1}{r}\) | Straight line with negative slope |
The correct evaluation is
ⓐ. P and Q are correct only
ⓑ. P, Q, and R are correct
ⓒ. P is correct only
ⓓ. Q and R are correct only
Correct Answer: P, Q, and R are correct
Explanation: For fixed masses, \(F=Gm_1m_2\left(\frac{1}{r^2}\right)\), so \(F\) versus \(\frac{1}{r^2}\) is a straight line through the origin. For circular orbits around the same central mass, \(T^2=\frac{4\pi^2}{GM}r^3\), so \(T^2\) versus \(r^3\) is also a straight line through the origin. For potential, \(V=-GM\left(\frac{1}{r}\right)\), so \(V\) versus \(\frac{1}{r}\) is a straight line with slope \(-GM\). The signs and axes must be read carefully because the three straight-line graphs represent different physical quantities.
316. A line graph of \(T^2\) against \(r^3\) for satellites around a planet has a smaller slope than the same type of graph for satellites around another planet. The first planet must have
ⓐ. satellites of larger mass only
ⓑ. zero gravitational field
ⓒ. a smaller central mass
ⓓ. a larger central mass
Correct Answer: a larger central mass
Explanation: For circular satellite motion, \(T^2=\frac{4\pi^2}{GM}r^3\). The slope of a \(T^2\) versus \(r^3\) graph is \(\frac{4\pi^2}{GM}\). If the slope is smaller, the denominator \(GM\) must be larger. Since \(G\) is universal, this means the central mass \(M\) is larger. The masses of the satellites do not affect this slope because satellite mass cancels in the orbital derivation.
317. A gravitational potential energy graph for two masses is compared with a gravitational force graph. The potential energy varies as \(-\frac{1}{r}\), while the force magnitude varies as
ⓐ. \(\frac{1}{r}\)
ⓑ. \(\frac{1}{\sqrt{r}}\)
ⓒ. \(\sqrt{r}\)
ⓓ. \(\frac{1}{r^2}\)
Correct Answer: \(\frac{1}{r^2}\)
Explanation: Gravitational potential energy of two masses is \(U=-G\frac{Mm}{r}\). This inverse-distance form comes from the work done against the inverse-square gravitational force. The force magnitude itself is \(F=G\frac{Mm}{r^2}\). Thus potential energy and force do not have the same distance power. The negative sign belongs to potential energy with zero at infinity, while force magnitude is written positive with direction handled separately.
318. A student sees a graph of \(U\) against \(r\) below the axis and says, “The force must be repulsive because the energy is negative.” The best correction is that
ⓐ. negative \(U\) means the interacting masses are negative
ⓑ. gravity becomes repulsive at large separation from the source
ⓒ. negative \(U\) means a bound state below the infinity zero
ⓓ. potential energy sign is arbitrary and has no meaning
Correct Answer: negative \(U\) means a bound state below the infinity zero
Explanation: Gravitational potential energy is negative for two attracting masses when \(U=0\) is chosen at infinity. The negative value means energy must be supplied to separate the masses to infinity without leftover kinetic energy. It does not mean the force is repulsive. The gravitational force between ordinary masses remains attractive and is directed along the line joining the masses. The sign of potential energy must always be interpreted with its reference level.
319. A graph of orbital speed \(v_o\) against orbital radius \(r\) for circular satellites is decreasing. If the horizontal axis is changed to \(\frac{1}{r}\), the relation between \(v_o^2\) and \(\frac{1}{r}\) becomes
ⓐ. a straight line through the origin
ⓑ. a straight line not through the origin
ⓒ. a line with negative slope
ⓓ. a horizontal line above the axis
Correct Answer: a straight line through the origin
Explanation: For a circular satellite, \(v_o=\sqrt{\frac{GM}{r}}\). Squaring both sides gives \(v_o^2=\frac{GM}{r}\). If \(x=\frac{1}{r}\), then \(v_o^2=GMx\). This is a straight-line relation through the origin with positive slope \(GM\). Plotting \(v_o\) directly against \(r\) does not give a straight line because of the square-root dependence.
320. A set of graph observations says: \(F\) falls by a factor of \(9\) when \(r\) triples, \(V\) becomes one-third when \(r\) triples, and \(T\) becomes \(3\sqrt{3}\) times when \(r\) triples for circular orbits around the same mass. The record is
ⓐ. only potential and period observations are correct
ⓑ. all three observations are correct
ⓒ. none of the three observations is correct
ⓓ. only force and potential observations are correct
Correct Answer: all three observations are correct
Explanation: \( \textbf{Force dependence:} \)
\[
F\propto\frac{1}{r^2}
\]
If \(r\) triples, then
\[
F'=\frac{F}{3^2}=\frac{F}{9}
\]
\( \textbf{Potential dependence:} \)
\[
V\propto-\frac{1}{r}
\]
Tripling \(r\) makes the potential one-third of its previous value, meaning it is less negative and closer to zero.
\( \textbf{Period dependence:} \)
\[
T\propto r^{3/2}
\]
So,
\[
\frac{T'}{T}=3^{3/2}=3\sqrt{3}
\]
\( \textbf{Final answer:} \) all three observations are correct, but they use three different powers of the same distance factor.
