401. A satellite moves in a circular orbit of radius \(2.5\times10^7\,\text{m}\) with speed \(4.0\,\text{km s}^{-1}\). Take \(G=6.67\times10^{-11}\,\text{N m}^2\text{kg}^{-2}\) and \(\pi=3.14\). What are the central mass and orbital period?
ⓐ. \(M\approx6.0\times10^{24}\,\text{kg},\quad T\approx7.85\times10^4\,\text{s}\)
ⓑ. \(M\approx3.0\times10^{24}\,\text{kg},\quad T\approx1.96\times10^4\,\text{s}\)
ⓒ. \(M\approx6.0\times10^{24}\,\text{kg},\quad T\approx3.93\times10^4\,\text{s}\)
ⓓ. \(M\approx1.2\times10^{25}\,\text{kg},\quad T\approx3.93\times10^4\,\text{s}\)
Correct Answer: \(M\approx6.0\times10^{24}\,\text{kg},\quad T\approx3.93\times10^4\,\text{s}\)
Explanation: \(\textbf{Infer the central mass from circular orbital speed:}\)
\[
v_o^2=\frac{GM}{r}
\]
Therefore,
\[
M=\frac{v_o^2r}{G}
\]
Convert the speed:
\[
v_o=4.0\times10^3\,\text{m s}^{-1}
\]
Then,
\[
M
=
\frac{(4.0\times10^3)^2(2.5\times10^7)}
{6.67\times10^{-11}}
\]
The numerator is
\[
(1.6\times10^7)(2.5\times10^7)
=4.0\times10^{14}
\]
Hence,
\[
M\approx6.0\times10^{24}\,\text{kg}
\]
\(\textbf{Use the orbital circumference to obtain the period:}\)
\[
T=\frac{2\pi r}{v_o}
\]
Substituting,
\[
T
=
\frac{2(3.14)(2.5\times10^7)}
{4.0\times10^3}
\]
Thus,
\[
T\approx3.93\times10^4\,\text{s}
\]
\(\textbf{Final answer:}\)
\[
\boxed{M\approx6.0\times10^{24}\,\text{kg},\qquad
T\approx3.93\times10^4\,\text{s}}
\]
402. Uniform spherical planets A and B have
\[
\rho_B=4\rho_A,\qquad R_B=\frac{R_A}{2}.
\]
A satellite orbits each planet at an altitude equal to that planet's own radius. If the corresponding orbital speeds and periods are \(v_A,v_B\) and \(T_A,T_B\), which pair is correct?
ⓐ. \(v_B/v_A=2,\quad T_B/T_A=1/4\)
ⓑ. \(v_B/v_A=1,\quad T_B/T_A=1/2\)
ⓒ. \(v_B/v_A=1/2,\quad T_B/T_A=1\)
ⓓ. \(v_B/v_A=\sqrt2,\quad T_B/T_A=1/\sqrt2\)
Correct Answer: \(v_B/v_A=1,\quad T_B/T_A=1/2\)
Explanation: \(\textbf{Compare the planetary masses:}\)
For a uniform sphere,
\[
M\propto\rho R^3
\]
Therefore,
\[
\frac{M_B}{M_A}
=
4\left(\frac12\right)^3
=
\frac12
\]
\(\textbf{Write the two orbital radii:}\)
Each satellite is at altitude equal to its planet's radius.
Thus,
\[
r_A=2R_A
\]
Also,
\[
r_B=2R_B=R_A
\]
Hence,
\[
\frac{r_B}{r_A}=\frac12
\]
\(\textbf{Compare circular orbital speeds:}\)
\[
v_o^2=\frac{GM}{r}
\]
Thus,
\[
\frac{v_B^2}{v_A^2}
=
\frac{M_B/r_B}{M_A/r_A}
\]
Substituting the ratios,
\[
\frac{v_B^2}{v_A^2}
=
\frac{(1/2)}{(1/2)}
=1
\]
Therefore,
\[
v_B=v_A
\]
\(\textbf{Compare the periods:}\)
Since
\[
T=\frac{2\pi r}{v},
\]
and the speeds are equal,
\[
\frac{T_B}{T_A}
=
\frac{r_B}{r_A}
=
\frac12
\]
\(\textbf{Final answer:}\)
\[
\boxed{\frac{v_B}{v_A}=1,\qquad
\frac{T_B}{T_A}=\frac12}
\]
403. The surface escape speed of a spherical planet is \(v_e\). A satellite moves in a circular orbit around the planet with speed \(v_e/2\). If the planetary radius is \(R\), what are the satellite's altitude and its period relative to the period \(T_0\) of a circular satellite just above the surface?
ⓐ. \(h=R,\quad T/T_0=2\sqrt2\)
ⓑ. \(h=2R,\quad T/T_0=3\sqrt3\)
ⓒ. \(h=R,\quad T/T_0=2\)
ⓓ. \(h=3R,\quad T/T_0=8\)
Correct Answer: \(h=R,\quad T/T_0=2\sqrt2\)
Explanation: \(\textbf{Relate the remote orbital speed to the surface escape speed:}\)
The surface escape speed satisfies
\[
v_e^2=\frac{2GM}{R}
\]
The satellite speed is
\[
v_o=\frac{v_e}{2}
\]
Therefore,
\[
v_o^2=\frac{v_e^2}{4}
=\frac{GM}{2R}
\]
For a circular orbit,
\[
v_o^2=\frac{GM}{r}
\]
Hence,
\[
\frac{GM}{r}
=
\frac{GM}{2R}
\]
Therefore,
\[
r=2R
\]
Thus,
\[
h=r-R=R
\]
\(\textbf{Compare with the near-surface circular period:}\)
For a fixed central mass,
\[
T\propto r^{3/2}
\]
Hence,
\[
\frac{T}{T_0}
=
\left(\frac{2R}{R}\right)^{3/2}
\]
Therefore,
\[
\frac{T}{T_0}=2\sqrt2
\]
\(\textbf{Final answer:}\)
\[
\boxed{h=R,\qquad \frac{T}{T_0}=2\sqrt2}
\]
404. For circular satellites around central bodies A and B, graphs of orbital speed \(v_o\) against \(1/\sqrt r\) are drawn.
Both graphs are straight lines through the origin. The slope of the graph for central body B is twice the slope of the graph for central body A.
For satellites at the same orbital radius around A and B, what are \(M_B/M_A\) and \(T_B/T_A\)?
ⓐ. \(2,\ 1/2\)
ⓑ. \(4,\ 1/2\)
ⓒ. \(4,\ 1/4\)
ⓓ. \(2,\ 1/\sqrt2\)
Correct Answer: \(4,\ 1/2\)
Explanation: \(\textbf{Interpret the speed-graph slope:}\)
For circular motion,
\[
v_o=\sqrt{\frac{GM}{r}}
\]
Therefore,
\[
v_o=\sqrt{GM}\left(\frac1{\sqrt r}\right)
\]
The graph slope is
\[
k=\sqrt{GM}
\]
Given
\[
k_B=2k_A,
\]
we have
\[
\sqrt{GM_B}
=
2\sqrt{GM_A}
\]
Squaring,
\[
M_B=4M_A
\]
\(\textbf{Compare periods at the same orbital radius:}\)
At fixed \(r\),
\[
T=2\pi\sqrt{\frac{r^3}{GM}}
\]
Therefore,
\[
T\propto\frac1{\sqrt M}
\]
Hence,
\[
\frac{T_B}{T_A}
=
\sqrt{\frac{M_A}{M_B}}
=
\frac12
\]
The larger central mass gives a steeper orbital-speed graph and a shorter period at the same radius.
\(\textbf{Final answer:}\)
\[
\boxed{\frac{M_B}{M_A}=4,\qquad
\frac{T_B}{T_A}=\frac12}
\]
405. Consider the following statements about ideal circular satellite motion.
Statement I: At fixed central mass, orbital speed decreases as orbital radius increases.
Statement II: Satellite mass does not appear in the ideal expression \(v_o=\sqrt{GM/r}\).
Statement III: The same expression \(v=\sqrt{GM/r}\) may be used as the actual instantaneous speed at every point of an arbitrary elliptical satellite orbit.
Statement IV: For a circular orbit just above a spherical planet's surface, \(v_o=\sqrt{gR}\).
Statement V: Two satellites of different masses in the same circular orbit around the same planet have the same orbital speed.
Which statements are correct?
ⓐ. I, II and III only
ⓑ. I, II, IV and V only
ⓒ. II, III, IV and V only
ⓓ. I, III and V only
Correct Answer: I, II, IV and V only
Explanation: Statement I is correct because
\[
v_o=\sqrt{\frac{GM}{r}},
\]
so at fixed \(M\),
\[
v_o\propto r^{-1/2}.
\]
Statement II is correct because satellite mass cancels when gravitational force is equated to the required centripetal force.
Statement III is incorrect.
The expression
\[
v_o=\sqrt{\frac{GM}{r}}
\]
is the circular-orbit speed at radius \(r\); an elliptical orbit must not be treated point-by-point as a set of independent circular orbits.
Statement IV is correct because at the planetary surface
\[
GM=gR^2.
\]
Hence,
\[
v_o=\sqrt{\frac{GM}{R}}=\sqrt{gR}.
\]
Statement V is correct because ideal orbital speed at fixed \(M\) and \(r\) is independent of satellite mass.
Therefore Statements I, II, IV and V only are correct.
406. For a satellite in a circular orbit of radius \(r\), which sequence correctly leads to its time period?
ⓐ. \(T=2\pi r/v_o\), hence \(T=2\pi\sqrt{GM/r^3}\)
ⓑ. \(T=v_o/(2\pi r)\), \(v_o=\sqrt{GM/r}\), hence \(T=\sqrt{GM/r^3}/(2\pi)\)
ⓒ. \(T=2\pi r\,v_o\), hence \(T=2\pi\sqrt{GMr}\)
ⓓ. \(T=2\pi r/v_o\), \(v_o=\sqrt{GM/r}\), hence \(T=2\pi\sqrt{r^3/(GM)}\)
Correct Answer: \(T=2\pi r/v_o\), \(v_o=\sqrt{GM/r}\), hence \(T=2\pi\sqrt{r^3/(GM)}\)
Explanation: For one complete circular revolution, the satellite travels the orbital circumference
\[
2\pi r.
\]
If its constant orbital speed is \(v_o\), then
\[
T=\frac{2\pi r}{v_o}.
\]
For gravitational circular motion,
\[
v_o=\sqrt{\frac{GM}{r}}.
\]
Substituting,
\[
T
=
\frac{2\pi r}{\sqrt{GM/r}}.
\]
Therefore,
\[
T
=
2\pi r\sqrt{\frac{r}{GM}}.
\]
Hence,
\[
T=2\pi\sqrt{\frac{r^3}{GM}}.
\]
Squaring gives
\[
T^2=\frac{4\pi^2r^3}{GM}.
\]
The satellite mass does not appear because it has already cancelled from the circular-orbit dynamics.
\(\textbf{Final answer:}\)
\[
T=2\pi\sqrt{\frac{r^3}{GM}}.
\]
407. A satellite moves in a circular orbit at altitude \(R\) above a spherical planet of radius \(R\) and surface gravity \(g\). What is its orbital period?
ⓐ. \(2\pi\sqrt{\dfrac{R}{g}}\)
ⓑ. \(2\pi\sqrt{\dfrac{2R}{g}}\)
ⓒ. \(4\pi\sqrt{\dfrac{2R}{g}}\)
ⓓ. \(8\pi\sqrt{\dfrac{R}{g}}\)
Correct Answer: \(4\pi\sqrt{\dfrac{2R}{g}}\)
Explanation: \(\textbf{Write the actual orbital radius:}\)
The altitude is
\[
h=R
\]
Therefore,
\[
r=R+h=2R
\]
\(\textbf{Express }GM\textbf{ using surface gravity:}\)
\[
g=\frac{GM}{R^2}
\]
Hence,
\[
GM=gR^2
\]
\(\textbf{Substitute into the satellite-period relation:}\)
\[
T=2\pi\sqrt{\frac{r^3}{GM}}
\]
Thus,
\[
T
=
2\pi
\sqrt{
\frac{(2R)^3}{gR^2}
}
\]
Therefore,
\[
T
=
2\pi\sqrt{\frac{8R}{g}}
\]
Since
\[
\sqrt8=2\sqrt2,
\]
\[
T=4\pi\sqrt{\frac{2R}{g}}
\]
Using altitude directly as \(r\) would miss the required centre-distance conversion.
\(\textbf{Final answer:}\)
\[
\boxed{T=4\pi\sqrt{\frac{2R}{g}}}
\]
408. A satellite just above a planet's surface has period \(T_0\) and orbital speed \(v_0\). Another circular satellite around the same planet has period \(27T_0\). What are its altitude and orbital speed?
ⓐ. \(h=8R,\quad v=v_0/3\)
ⓑ. \(h=9R,\quad v=v_0/9\)
ⓒ. \(h=8R,\quad v=v_0/9\)
ⓓ. \(h=26R,\quad v=v_0/3\)
Correct Answer: \(h=8R,\quad v=v_0/3\)
Explanation: \(\textbf{Infer orbital radius from the period ratio:}\)
For the same planet,
\[
T\propto r^{3/2}
\]
The near-surface satellite has
\[
r_0=R
\]
Therefore,
\[
\frac{T}{T_0}
=
\left(\frac rR\right)^{3/2}
\]
Given
\[
\frac{T}{T_0}=27,
\]
we obtain
\[
27=\left(\frac rR\right)^{3/2}
\]
Raising both sides to the power \(2/3\),
\[
\frac rR=9
\]
Thus,
\[
r=9R
\]
The altitude is therefore
\[
h=r-R=8R
\]
\(\textbf{Use the speed-radius dependence:}\)
For circular motion around the same planet,
\[
v\propto r^{-1/2}
\]
Hence,
\[
\frac v{v_0}
=
\sqrt{\frac{R}{9R}}
=\frac13
\]
Therefore,
\[
v=\frac{v_0}{3}
\]
\(\textbf{Final answer:}\)
\[
\boxed{h=8R,\qquad v=\frac{v_0}{3}}
\]
409. Circular satellites A and B revolve around the same planet. Their periods satisfy
\[
T_B=8T_A.
\]
What are the ratios of their local gravitational accelerations and orbital speeds?
ⓐ. \(g_B/g_A=1/8,\quad v_B/v_A=1/2\)
ⓑ. \(g_B/g_A=1/16,\quad v_B/v_A=1/4\)
ⓒ. \(g_B/g_A=1/16,\quad v_B/v_A=1/2\)
ⓓ. \(g_B/g_A=1/64,\quad v_B/v_A=1/4\)
Correct Answer: \(g_B/g_A=1/16,\quad v_B/v_A=1/2\)
Explanation: \(\textbf{Translate the period ratio into a radius ratio:}\)
For the same central mass,
\[
T\propto r^{3/2}
\]
Therefore,
\[
\frac{T_B}{T_A}
=
\left(\frac{r_B}{r_A}\right)^{3/2}
\]
Given
\[
\frac{T_B}{T_A}=8,
\]
we obtain
\[
\frac{r_B}{r_A}=8^{2/3}=4
\]
\(\textbf{Compare the local gravitational accelerations:}\)
\[
g_r\propto\frac1{r^2}
\]
Hence,
\[
\frac{g_B}{g_A}
=
\left(\frac{r_A}{r_B}\right)^2
=
\frac1{16}
\]
\(\textbf{Compare the orbital speeds:}\)
\[
v_o\propto r^{-1/2}
\]
Therefore,
\[
\frac{v_B}{v_A}
=
\sqrt{\frac{r_A}{r_B}}
=
\frac12
\]
A period increase by a factor \(8\) therefore corresponds to a fourfold orbital-radius increase, a sixteenfold decrease in local \(g\), and a twofold decrease in orbital speed.
\(\textbf{Final answer:}\)
\[
\boxed{\frac{g_B}{g_A}=\frac1{16},\qquad
\frac{v_B}{v_A}=\frac12}
\]
410. A spherical planet has radius \(R\) and surface escape speed \(v_e\). A circular satellite around it has period
\[
T=\frac{8\pi R}{v_e}.
\]
What are the satellite's altitude and orbital speed?
ⓐ. \(h=R,\quad v_o=v_e/2\)
ⓑ. \(h=2R,\quad v_o=v_e/\sqrt6\)
ⓒ. \(h=R,\quad v_o=v_e/\sqrt2\)
ⓓ. \(h=3R,\quad v_o=v_e/4\)
Correct Answer: \(h=R,\quad v_o=v_e/2\)
Explanation: \(\textbf{Express the central mass parameter using surface escape speed:}\)
At the surface,
\[
v_e^2=\frac{2GM}{R}
\]
Therefore,
\[
GM=\frac{v_e^2R}{2}
\]
\(\textbf{Insert this into the satellite-period relation:}\)
\[
T=2\pi\sqrt{\frac{r^3}{GM}}
\]
Thus,
\[
T
=
2\pi
\sqrt{
\frac{2r^3}{v_e^2R}
}
\]
The given period is
\[
\frac{8\pi R}{v_e}
\]
Hence,
\[
\frac{8\pi R}{v_e}
=
\frac{2\pi}{v_e}
\sqrt{\frac{2r^3}{R}}
\]
Canceling \(2\pi/v_e\),
\[
4R=\sqrt{\frac{2r^3}{R}}
\]
Squaring,
\[
16R^2=\frac{2r^3}{R}
\]
Therefore,
\[
r^3=8R^3,
\]
so
\[
r=2R
\]
Hence,
\[
h=r-R=R
\]
\(\textbf{Find the circular orbital speed:}\)
\[
v_o^2=\frac{GM}{r}
\]
Substituting,
\[
v_o^2
=
\frac{v_e^2R/2}{2R}
=
\frac{v_e^2}{4}
\]
Thus,
\[
v_o=\frac{v_e}{2}
\]
\(\textbf{Final answer:}\)
\[
\boxed{h=R,\qquad v_o=\frac{v_e}{2}}
\]
411. At centre-distance \(r\) from a spherical planet, a satellite has speed equal in magnitude to the circular-orbit speed \(v_c=\sqrt{GM/r}\), but its velocity makes an angle \(60^\circ\) with the outward radius vector. Which conclusion is correct?
ⓐ. It is circular because its speed magnitude equals \(v_c\)
ⓑ. It is non-circular but bound, with total energy \(-GMm/(2r)\)
ⓒ. It is at the escape threshold because its velocity has a radial component
ⓓ. It is unbound and must reach infinity because its speed magnitude equals \(v_c\)
Correct Answer: It is non-circular but bound, with total energy \(-GMm/(2r)\)
Explanation: A circular orbit requires more than the correct speed magnitude.
The instantaneous velocity must also be tangential to the circular path and therefore perpendicular to the radius vector.
Here the velocity makes an angle \(60^\circ\) with the radius, so it has a non-zero radial component.
Hence the instantaneous state cannot represent a circular orbit.
However, the kinetic energy depends only on the speed magnitude:
\[
K=\frac12mv_c^2
\]
Using
\[
v_c^2=\frac{GM}{r},
\]
\[
K=\frac{GMm}{2r}
\]
The potential energy at the same point is
\[
U=-\frac{GMm}{r}
\]
Therefore,
\[
E=K+U
=
-\frac{GMm}{2r}
\]
The negative total energy shows that the satellite is bound even though the orbit is not circular.
\(\textbf{Final answer:}\) The satellite is in a bound non-circular orbit with
\[
E=-\frac{GMm}{2r}
\]
412. A satellite at centre-distance \(r\) is given a purely tangential speed \(v_c/2\), where
\[
v_c=\sqrt{\frac{GM}{r}}
\]
is the circular-orbit speed at that radius. Which pair correctly gives the ratio of the actual gravitational acceleration to the centripetal acceleration required for a circle of radius \(r\) at the given speed, and the satellite's total mechanical energy at that instant?
ⓐ. \(1,\ -GMm/(2r)\)
ⓑ. \(2,\ -3GMm/(4r)\)
ⓒ. \(4,\ -7GMm/(8r)\)
ⓓ. \(4,\ 0\)
Correct Answer: \(4,\ -7GMm/(8r)\)
Explanation: \(\textbf{Compare the actual and required inward accelerations:}\)
The actual gravitational acceleration is
\[
a_g=\frac{GM}{r^2}
\]
Since
\[
v_c^2=\frac{GM}{r},
\]
this may be written as
\[
a_g=\frac{v_c^2}{r}
\]
At the given speed \(v=v_c/2\), the inward acceleration required to maintain a circle of the same radius would be
\[
a_{\text{req}}=\frac{v^2}{r}
\]
Thus,
\[
a_{\text{req}}
=
\frac{v_c^2}{4r}
\]
Therefore,
\[
\frac{a_g}{a_{\text{req}}}=4
\]
Gravity is too strong for that speed to maintain the original circular radius, so the orbit cannot remain that circle.
\(\textbf{Check the signed orbital energy:}\)
The kinetic energy is
\[
K=\frac12m\left(\frac{v_c}{2}\right)^2
=\frac18mv_c^2
\]
Hence,
\[
K=\frac{GMm}{8r}
\]
With
\[
U=-\frac{GMm}{r},
\]
\[
E
=
\frac{GMm}{8r}
-\frac{GMm}{r}
=
-\frac{7GMm}{8r}
\]
The negative energy confirms that the satellite remains gravitationally bound.
\(\textbf{Final answer:}\)
\[
\boxed{\frac{a_g}{a_{\text{req}}}=4,\qquad
E=-\frac{7GMm}{8r}}
\]
413. A satellite moves in a circular orbit of radius \(r\) with speed \(v_c\). At one point, an ideal instantaneous manoeuvre reverses its velocity direction exactly while leaving its speed magnitude unchanged. Immediately after the manoeuvre, which statement is correct?
ⓐ. It remains on the same circular orbit in reverse, with unchanged period and total energy
ⓑ. It enters an escape trajectory because the reversal gives a large momentum change
ⓒ. It falls radially inward because its angular momentum becomes zero
ⓓ. It enters a larger ellipse because its kinetic energy increases
Correct Answer: It remains on the same circular orbit in reverse, with unchanged period and total energy
Explanation: Before the manoeuvre, the satellite velocity is tangential and has magnitude
\[
v_c=\sqrt{\frac{GM}{r}}.
\]
After exact reversal, the new velocity is still tangential to the same circle.
Its magnitude is still \(v_c\).
Therefore the required centripetal acceleration remains
\[
\frac{v_c^2}{r}.
\]
Gravity at that position still provides
\[
\frac{GM}{r^2}
=
\frac{v_c^2}{r}.
\]
Thus the circular-orbit condition remains satisfied.
The angular momentum changes sign because the direction of motion reverses, but its magnitude does not vanish.
The kinetic energy is unchanged because the speed magnitude is unchanged.
The potential energy is also unchanged because the radius is unchanged.
Hence the total energy and orbital period remain the same, while the direction of revolution reverses.
414. A sealed capsule is falling freely toward Earth along a purely radial trajectory, and an astronaut inside it is not touching the walls. Neglect air resistance and tidal differences across the capsule. Which statement is correct?
ⓐ. The astronaut cannot be weightless because weightlessness occurs only in circular orbit
ⓑ. The astronaut can be weightless because both astronaut and capsule share essentially the same free-fall acceleration
ⓒ. The astronaut is weightless only because Earth's gravitational field inside the capsule becomes zero
ⓓ. The astronaut must experience a normal force equal to \(mg\) because the capsule is accelerating
Correct Answer: The astronaut can be weightless because both astronaut and capsule share essentially the same free-fall acceleration
Explanation: Weightlessness is fundamentally a free-fall condition rather than a property unique to circular orbital motion.
Earth's gravitational force continues to act on both the astronaut and the capsule.
Both undergo essentially the same gravitational acceleration under the stated idealisation.
Therefore the capsule floor does not need to provide a sustained supporting normal force on the astronaut.
Apparent weight is associated with this support force.
Hence the apparent weight can be zero even during purely radial free fall.
The astronaut's gravitational mass and gravitational force have not vanished.
An orbiting spacecraft is another example of continuous free fall, but orbital motion is not required for the basic weightlessness condition.
Thus option B correctly identifies the physical mechanism.
415. A circular satellite has orbital period
\[
T=2000\pi\,\text{s}
\]
at a location where the gravitational acceleration is
\[
4.0\,\text{m s}^{-2}.
\]
What are its orbital radius, orbital speed and the central gravitational parameter \(GM\)?
ⓐ. \(r=8.0\times10^6\,\text{m},\ v=4.0\,\text{km s}^{-1},\ GM=1.28\times10^{14}\,\text{m}^3\text{s}^{-2}\)
ⓑ. \(r=2.0\times10^6\,\text{m},\ v=2.0\,\text{km s}^{-1},\ GM=1.6\times10^{13}\,\text{m}^3\text{s}^{-2}\)
ⓒ. \(r=4.0\times10^6\,\text{m},\ v=2.0\,\text{km s}^{-1},\ GM=3.2\times10^{13}\,\text{m}^3\text{s}^{-2}\)
ⓓ. \(r=4.0\times10^6\,\text{m},\ v=4.0\,\text{km s}^{-1},\ GM=6.4\times10^{13}\,\text{m}^3\text{s}^{-2}\)
Correct Answer: \(r=4.0\times10^6\,\text{m},\ v=4.0\,\text{km s}^{-1},\ GM=6.4\times10^{13}\,\text{m}^3\text{s}^{-2}\)
Explanation: \(\textbf{Combine circular acceleration with the measured period:}\)
For circular motion,
\[
g_r=\frac{4\pi^2r}{T^2}
\]
Therefore,
\[
r=\frac{g_rT^2}{4\pi^2}
\]
Substituting,
\[
r
=
\frac{4(2000\pi)^2}{4\pi^2}
\]
Hence,
\[
r=4.0\times10^6\,\text{m}
\]
\(\textbf{Use the circumference to obtain orbital speed:}\)
\[
v=\frac{2\pi r}{T}
\]
Thus,
\[
v
=
\frac{2\pi(4.0\times10^6)}
{2000\pi}
\]
Therefore,
\[
v=4.0\times10^3\,\text{m s}^{-1}
=4.0\,\text{km s}^{-1}
\]
\(\textbf{Recover the central gravitational parameter:}\)
\[
g_r=\frac{GM}{r^2}
\]
Hence,
\[
GM=g_rr^2
\]
Substitution gives
\[
GM
=
4(4.0\times10^6)^2
=6.4\times10^{13}\,\text{m}^3\text{s}^{-2}
\]
\(\textbf{Final answer:}\)
\[
\boxed{r=4.0\times10^6\,\text{m},\quad
v=4.0\,\text{km s}^{-1},\quad
GM=6.4\times10^{13}\,\text{m}^3\text{s}^{-2}}
\]
416. For circular satellites around one central body, \(T^2\) is plotted against \(r^3\).
The graph is a straight line through the origin with slope \(S\). A satellite of interest moves at orbital radius \(r_0\).
Which pair correctly gives the central mass \(M\) and the orbital speed \(v_0\) of that satellite?
ⓐ. \(M=\dfrac{4\pi^2}{GS},\quad v_0=\dfrac{2\pi}{\sqrt{Sr_0}}\)
ⓑ. \(M=\dfrac{S}{4\pi^2G},\quad v_0=2\pi\sqrt{Sr_0}\)
ⓒ. \(M=\dfrac{4\pi^2S}{G},\quad v_0=\dfrac{2\pi}{\sqrt{Sr_0}}\)
ⓓ. \(M=\dfrac{4\pi^2}{GS},\quad v_0=2\pi\sqrt{\dfrac{r_0}{S}}\)
Correct Answer: \(M=\dfrac{4\pi^2}{GS},\quad v_0=\dfrac{2\pi}{\sqrt{Sr_0}}\)
Explanation: \(\textbf{Interpret the period graph:}\)
For circular satellites,
\[
T^2=\frac{4\pi^2r^3}{GM}
\]
Therefore the slope of a graph of \(T^2\) against \(r^3\) is
\[
S=\frac{4\pi^2}{GM}
\]
Hence,
\[
GM=\frac{4\pi^2}{S}
\]
Thus,
\[
M=\frac{4\pi^2}{GS}
\]
\(\textbf{Use the inferred gravitational parameter for orbital speed:}\)
At radius \(r_0\),
\[
v_0^2=\frac{GM}{r_0}
\]
Substitution gives
\[
v_0^2
=
\frac{4\pi^2}{Sr_0}
\]
Therefore,
\[
v_0=\frac{2\pi}{\sqrt{Sr_0}}
\]
A smaller period-graph slope corresponds to a larger central mass and therefore to a larger circular speed at a fixed radius.
\(\textbf{Final answer:}\)
\[
\boxed{M=\frac{4\pi^2}{GS},\qquad
v_0=\frac{2\pi}{\sqrt{Sr_0}}}
\]
417. For circular satellites around a fixed central body, \(Y=v_o^3\) is plotted against
\[
X=\frac1T.
\]
The graph is a straight line through the origin with slope \(S\).
Which relation correctly follows from the graph?
ⓐ. \(M=\dfrac{S}{2\pi G}\), and at \(X=x_0\), \(v_o=(Sx_0)^{1/3}\)
ⓑ. \(M=\dfrac{2\pi S}{G}\), and at \(X=x_0\), \(v_o=Sx_0\)
ⓒ. \(M=\dfrac{S}{4\pi^2G}\), and at \(X=x_0\), \(v_o=\sqrt{Sx_0}\)
ⓓ. \(M=\dfrac{1}{2\pi GS}\), and at \(X=x_0\), \(v_o=(x_0/S)^{1/3}\)
Correct Answer: \(M=\dfrac{S}{2\pi G}\), and at \(X=x_0\), \(v_o=(Sx_0)^{1/3}\)
Explanation: \(\textbf{Eliminate orbital radius between speed and period:}\)
For a circular satellite,
\[
v_o^2=\frac{GM}{r}
\]
Also,
\[
T=\frac{2\pi r}{v_o}
\]
Hence,
\[
r=\frac{v_oT}{2\pi}
\]
Substituting this into the speed relation,
\[
v_o^2
=
\frac{GM}{v_oT/(2\pi)}
\]
Therefore,
\[
v_o^3T=2\pi GM
\]
Thus,
\[
v_o^3
=
2\pi GM\left(\frac1T\right)
\]
With
\[
Y=v_o^3,\qquad X=\frac1T,
\]
the graph slope is
\[
S=2\pi GM
\]
Hence,
\[
M=\frac{S}{2\pi G}
\]
At \(X=x_0\),
\[
v_o^3=Sx_0,
\]
so
\[
v_o=(Sx_0)^{1/3}
\]
\(\textbf{Final answer:}\)
\[
M=\frac{S}{2\pi G},\qquad v_o=(Sx_0)^{1/3}
\]
418. Four circular-satellite observations are claimed to correspond to the same central body. A reference satellite has orbital speed \(v\) and period \(T\).
| Observation | Orbital speed | Period |
|---|
| P | \(v\) | \(T\) |
| Q | \(2v\) | \(T/8\) |
| R | \(v/2\) | \(8T\) |
| S | \(3v\) | \(T/18\) |
Which observation is inconsistent?
ⓐ. P
ⓑ. Q
ⓒ. R
ⓓ. S
Correct Answer: S
Explanation: For circular satellites around the same central mass,
\[
v_o^3T=2\pi GM.
\]
Therefore the product
\[
v_o^3T
\]
must be the same for all four observations.
For P,
\[
v^3T
\]
is the reference value.
For Q,
\[
(2v)^3\left(\frac T8\right)
=
8v^3\frac T8
=
v^3T.
\]
Thus Q is consistent.
For R,
\[
\left(\frac v2\right)^3(8T)
=
\frac{v^3}{8}(8T)
=
v^3T.
\]
Thus R is also consistent.
For S,
\[
(3v)^3\left(\frac{T}{18}\right)
=
27v^3\frac{T}{18}
=
\frac32v^3T.
\]
Therefore S cannot correspond to the same central mass.
For speed \(3v\), the required period would instead be
\[
T_S=\frac{T}{27}.
\]
\(\textbf{Final answer:}\) Observation S is inconsistent.
419. Satellites A and B move in circular orbits around two different spherical central bodies. Their orbital periods are equal, while
\[
v_B=2v_A.
\]
What are
\[
\frac{r_B}{r_A},\qquad
\frac{M_B}{M_A},\qquad
\frac{g_B}{g_A}
\]
at the respective orbital radii?
ⓐ. \(2,\ 8,\ 2\)
ⓑ. \(4,\ 8,\ 1\)
ⓒ. \(2,\ 4,\ 2\)
ⓓ. \(4,\ 16,\ 4\)
Correct Answer: \(2,\ 8,\ 2\)
Explanation: \(\textbf{Use equal periods to relate speed and radius:}\)
For circular motion,
\[
v=\frac{2\pi r}{T}
\]
Since the two periods are equal,
\[
v\propto r
\]
Therefore,
\[
\frac{r_B}{r_A}
=
\frac{v_B}{v_A}=2
\]
\(\textbf{Infer the central-mass ratio:}\)
The circular-period relation is
\[
M=\frac{4\pi^2r^3}{GT^2}
\]
At equal \(T\),
\[
M\propto r^3
\]
Hence,
\[
\frac{M_B}{M_A}=2^3=8
\]
\(\textbf{Compare local gravitational accelerations:}\)
For circular motion,
\[
g_r=\frac{v^2}{r}
\]
Therefore,
\[
\frac{g_B}{g_A}
=
\frac{v_B^2}{v_A^2}
\frac{r_A}{r_B}
\]
Thus,
\[
\frac{g_B}{g_A}
=
4\left(\frac12\right)
=2
\]
\(\textbf{Final answer:}\)
\[
\boxed{\frac{r_B}{r_A}=2,\qquad
\frac{M_B}{M_A}=8,\qquad
\frac{g_B}{g_A}=2}
\]
420. Uniform spherical planets A and B have the same mean density, but
\[
R_B=3R_A.
\]
A satellite moves around each planet in a circular orbit whose centre-distance is twice that planet's radius. What are \(v_B/v_A\) and \(T_B/T_A\)?
ⓐ. \(1,\ 3\)
ⓑ. \(3,\ 1\)
ⓒ. \(3,\ 3\)
ⓓ. \(\sqrt3,\ 1/\sqrt3\)
Correct Answer: \(3,\ 1\)
Explanation: \(\textbf{Compare central masses at equal density:}\)
For uniform spheres,
\[
M\propto\rho R^3
\]
At equal density,
\[
\frac{M_B}{M_A}=3^3=27
\]
\(\textbf{Compare the satellite orbital radii:}\)
Each satellite has
\[
r=2R
\]
Therefore,
\[
\frac{r_B}{r_A}=3
\]
\(\textbf{Compare circular speeds:}\)
\[
v_o^2=\frac{GM}{r}
\]
Hence,
\[
\frac{v_B^2}{v_A^2}
=
\frac{M_B}{M_A}
\frac{r_A}{r_B}
\]
Thus,
\[
\frac{v_B^2}{v_A^2}
=
27\left(\frac13\right)=9
\]
Therefore,
\[
\frac{v_B}{v_A}=3
\]
\(\textbf{Compare orbital periods:}\)
\[
T^2\propto\frac{r^3}{M}
\]
Hence,
\[
\frac{T_B^2}{T_A^2}
=
\frac{3^3}{27}=1
\]
Therefore,
\[
\frac{T_B}{T_A}=1
\]
At the same density and same fractional orbital radius, planetary size changes the orbital speed but not the period.
\(\textbf{Final answer:}\)
\[
\boxed{\frac{v_B}{v_A}=3,\qquad
\frac{T_B}{T_A}=1}
\]
