201. Planet P revolves around star A with semi-major axis \(2r\) and period \(T\). Planet Q revolves around star B with semi-major axis \(4r\) and also has period \(T\). A third planet R revolves around star B with semi-major axis \(r\). Neglect planetary masses. What are \(M_B/M_A\) and \(T_R/T\)?
ⓐ. \(4,\ 1/4\)
ⓑ. \(8,\ 1/8\)
ⓒ. \(8,\ 1/4\)
ⓓ. \(16,\ 1/8\)
Correct Answer: \(8,\ 1/8\)
Explanation: \(\textbf{Compare the two central masses using equal periods:}\)
Newton's form of Kepler's third law gives
\[
T^2=\frac{4\pi^2a^3}{GM}
\]
For P and Q, the periods are equal:
\[
\frac{(2r)^3}{M_A}
=
\frac{(4r)^3}{M_B}
\]
Thus,
\[
\frac{8r^3}{M_A}
=
\frac{64r^3}{M_B}
\]
Therefore,
\[
M_B=8M_A
\]
\(\textbf{Now compare R with Q around the same star B:}\)
For the same central mass,
\[
T\propto a^{3/2}
\]
Hence,
\[
\frac{T_R}{T_Q}
=
\left(\frac{r}{4r}\right)^{3/2}
\]
Thus,
\[
\frac{T_R}{T}
=
\left(\frac14\right)^{3/2}
=\frac18
\]
The first comparison uses both orbital size and central mass, whereas the second reduces to the ordinary Kepler scaling because the central body is the same.
\(\textbf{Final answer:}\)
\[
\boxed{\frac{M_B}{M_A}=8,\qquad
\frac{T_R}{T}=\frac18}
\]
202. A small body moves in a circular gravitational orbit with constant speed \(v\) and period \(T\) around a central mass \(M\). Which pair correctly expresses the orbital radius \(r\) and central mass \(M\) using only \(v,T,G\)?
ⓐ. \(r=\dfrac{vT}{2\pi},\quad M=\dfrac{2\pi v^3T}{G}\)
ⓑ. \(r=\dfrac{2\pi v}{T},\quad M=\dfrac{v^2T}{2\pi G}\)
ⓒ. \(r=\dfrac{vT}{\pi},\quad M=\dfrac{v^3T}{4\pi G}\)
ⓓ. \(r=\dfrac{vT}{2\pi},\quad M=\dfrac{v^3T}{2\pi G}\)
Correct Answer: \(r=\dfrac{vT}{2\pi},\quad M=\dfrac{v^3T}{2\pi G}\)
Explanation: \(\textbf{Recover the orbital radius from the measured motion:}\)
For uniform circular motion,
\[
v=\frac{2\pi r}{T}
\]
Therefore,
\[
r=\frac{vT}{2\pi}
\]
\(\textbf{Use gravity as the centripetal interaction:}\)
For a circular gravitational orbit,
\[
\frac{GMm}{r^2}
=
\frac{mv^2}{r}
\]
The orbiting mass cancels:
\[
GM=v^2r
\]
Hence,
\[
M=\frac{v^2r}{G}
\]
Substitute the radius obtained from the kinematic relation:
\[
M
=
\frac{v^2}{G}
\left(\frac{vT}{2\pi}\right)
\]
Therefore,
\[
M=\frac{v^3T}{2\pi G}
\]
Thus the pair \(v,T\) is sufficient to recover both the orbital radius and the central mass under the stated circular-orbit assumptions.
\(\textbf{Final answer:}\)
\[
\boxed{r=\frac{vT}{2\pi},\qquad
M=\frac{v^3T}{2\pi G}}
\]
203. A small body is observed in a circular orbit of period \(T\). The gravitational acceleration produced by the central body at the orbital radius is measured to be \(g_o\). Which pair correctly gives the orbital radius and the central mass?
ⓐ. \(r=\dfrac{g_oT^2}{4\pi^2},\quad M=\dfrac{g_o^3T^4}{16\pi^4G}\)
ⓑ. \(r=\dfrac{4\pi^2g_o}{T^2},\quad M=\dfrac{g_oT^2}{4\pi^2G}\)
ⓒ. \(r=\dfrac{g_oT}{2\pi},\quad M=\dfrac{g_o^2T^2}{4\pi^2G}\)
ⓓ. \(r=\dfrac{g_oT^2}{2\pi},\quad M=\dfrac{g_o^3T^4}{4\pi^2G}\)
Correct Answer: \(r=\dfrac{g_oT^2}{4\pi^2},\quad M=\dfrac{g_o^3T^4}{16\pi^4G}\)
Explanation: \(\textbf{Identify the circular acceleration:}\)
At the orbital radius, gravity supplies the centripetal acceleration.
Therefore,
\[
g_o=\frac{v^2}{r}
\]
For a circular orbit,
\[
v=\frac{2\pi r}{T}
\]
Substitution gives
\[
g_o
=
\frac{(2\pi r/T)^2}{r}
\]
Hence,
\[
g_o=\frac{4\pi^2r}{T^2}
\]
Therefore,
\[
r=\frac{g_oT^2}{4\pi^2}
\]
\(\textbf{Recover the central mass:}\)
At the same radius,
\[
g_o=\frac{GM}{r^2}
\]
Thus,
\[
M=\frac{g_or^2}{G}
\]
Substituting the inferred radius,
\[
M=
\frac{g_o}{G}
\left(\frac{g_oT^2}{4\pi^2}\right)^2
\]
Therefore,
\[
M=\frac{g_o^3T^4}{16\pi^4G}
\]
\(\textbf{Final answer:}\)
\[
\boxed{r=\frac{g_oT^2}{4\pi^2},\qquad
M=\frac{g_o^3T^4}{16\pi^4G}}
\]
204. Use the graph comparison below.
For planets orbiting star A, a graph of \(T\) against \(a^{3/2}\) is a straight line through the origin with slope \(S\). For planets orbiting star B, the corresponding slope is \(S/3\).
If small bodies are placed in circular orbits of the same radius around the two stars, what are \(M_B/M_A\) and \(v_B/v_A\)?
ⓐ. \(3,\ \sqrt3\)
ⓑ. \(9,\ 3\)
ⓒ. \(9,\ 9\)
ⓓ. \(1/9,\ 1/3\)
Correct Answer: \(9,\ 3\)
Explanation: \(\textbf{Interpret the transformed Kepler graph:}\)
Newton's Kepler relation can be written as
\[
T
=
\frac{2\pi}{\sqrt{GM}}a^{3/2}
\]
Therefore the slope of a \(T\)-versus-\(a^{3/2}\) graph is
\[
S=\frac{2\pi}{\sqrt{GM}}
\]
Thus,
\[
S\propto\frac{1}{\sqrt M}
\]
Given
\[
S_B=\frac{S_A}{3},
\]
we have
\[
\frac{1}{\sqrt{M_B}}
=
\frac{1}{3\sqrt{M_A}}
\]
Hence,
\[
M_B=9M_A
\]
\(\textbf{Compare circular speeds at equal radius:}\)
\[
v=\sqrt{\frac{GM}{r}}
\]
At the same \(r\),
\[
\frac{v_B}{v_A}
=
\sqrt{\frac{M_B}{M_A}}
=\sqrt9=3
\]
\(\textbf{Final answer:}\)
\[
\boxed{\frac{M_B}{M_A}=9,\qquad
\frac{v_B}{v_A}=3}
\]
205. Star A is a uniform sphere of mean density \(\rho\) and radius \(R\). Star B is a uniform sphere of mean density \(4\rho\) and radius \(3R\). A small body moves in a circular orbit just outside the surface of each star. Neglect rotation and atmospheric effects. What are \(T_B/T_A\) and \(M_B/M_A\)?
ⓐ. \(2,\ 108\)
ⓑ. \(1/4,\ 27\)
ⓒ. \(1/2,\ 27\)
ⓓ. \(1/2,\ 108\)
Correct Answer: \(1/2,\ 108\)
Explanation: \(\textbf{Derive the near-surface orbital period in terms of density:}\)
For a circular orbit just outside a spherical body's surface,
\[
T^2=\frac{4\pi^2R^3}{GM}
\]
For a uniform sphere,
\[
M=\frac43\pi R^3\rho
\]
Substitution gives
\[
T^2
=
\frac{4\pi^2R^3}
{G(4\pi R^3\rho/3)}
\]
Therefore,
\[
T^2=\frac{3\pi}{G\rho}
\]
Thus the near-surface circular period is independent of radius for equal-density spherical bodies and satisfies
\[
T\propto\frac1{\sqrt\rho}
\]
Since
\[
\rho_B=4\rho_A,
\]
\[
\frac{T_B}{T_A}=\frac12
\]
\(\textbf{Compare the stellar masses:}\)
\[
M\propto\rho R^3
\]
Hence,
\[
\frac{M_B}{M_A}
=
4(3)^3
=108
\]
\(\textbf{Final answer:}\)
\[
\boxed{\frac{T_B}{T_A}=\frac12,\qquad
\frac{M_B}{M_A}=108}
\]
206. A planet moves around a star in an elliptical orbit with perihelion distance \(r\), aphelion distance \(3r\), and orbital period \(T\). What is the star's mass \(M\)? If another planet moves around the same star in a circular orbit of radius \(8r\), what is its period \(T_2\)?
ⓐ. \(M=\dfrac{8\pi^2r^3}{GT^2},\quad T_2=4T\)
ⓑ. \(M=\dfrac{32\pi^2r^3}{GT^2},\quad T_2=8T\)
ⓒ. \(M=\dfrac{32\pi^2r^3}{GT^2},\quad T_2=4T\)
ⓓ. \(M=\dfrac{16\pi^2r^3}{GT^2},\quad T_2=8T\)
Correct Answer: \(M=\dfrac{32\pi^2r^3}{GT^2},\quad T_2=8T\)
Explanation: \(\textbf{Find the semi-major axis of the elliptical orbit:}\)
\[
a=\frac{r_p+r_a}{2}
\]
Therefore,
\[
a=\frac{r+3r}{2}=2r
\]
\(\textbf{Infer the central mass from the orbital period:}\)
Newton's Kepler relation is
\[
T^2=\frac{4\pi^2a^3}{GM}
\]
Hence,
\[
M=\frac{4\pi^2a^3}{GT^2}
\]
Using
\[
a=2r,
\]
\[
M
=
\frac{4\pi^2(2r)^3}{GT^2}
\]
Therefore,
\[
M=\frac{32\pi^2r^3}{GT^2}
\]
\(\textbf{Compare the second orbit with the first:}\)
For the circular orbit,
\[
a_2=8r
\]
Both planets orbit the same star, so
\[
\frac{T_2}{T}
=
\left(\frac{8r}{2r}\right)^{3/2}
\]
Thus,
\[
\frac{T_2}{T}=4^{3/2}=8
\]
\(\textbf{Final answer:}\)
\[
\boxed{M=\frac{32\pi^2r^3}{GT^2},\qquad
T_2=8T}
\]
207. Planet A revolves around a star of mass \(M\) in an ellipse with perihelion distance \(r\) and aphelion distance \(7r\). Planet B revolves around another star of mass \(4M\) in an ellipse with perihelion distance \(2r\) and aphelion distance \(14r\). If their periods are \(T_A,T_B\), and B's perihelion and aphelion speeds are \(v_p,v_a\), then
ⓐ. \(\dfrac{T_B}{T_A}=2,\quad\dfrac{v_p}{v_a}=7\)
ⓑ. \(\dfrac{T_B}{T_A}=1,\quad\dfrac{v_p}{v_a}=7\)
ⓒ. \(\dfrac{T_B}{T_A}=\sqrt2,\quad\dfrac{v_p}{v_a}=\sqrt7\)
ⓓ. \(\dfrac{T_B}{T_A}=\sqrt2,\quad\dfrac{v_p}{v_a}=7\)
Correct Answer: \(\dfrac{T_B}{T_A}=\sqrt2,\quad\dfrac{v_p}{v_a}=7\)
Explanation: \(\textbf{Determine the two semi-major axes:}\)
For A,
\[
a_A=\frac{r+7r}{2}=4r
\]
For B,
\[
a_B=\frac{2r+14r}{2}=8r
\]
\(\textbf{Compare periods when the central masses differ:}\)
Newton's Kepler relation gives
\[
T^2\propto\frac{a^3}{M}
\]
Therefore,
\[
\left(\frac{T_B}{T_A}\right)^2
=
\left(\frac{a_B}{a_A}\right)^3
\frac{M}{4M}
\]
Substituting,
\[
\left(\frac{T_B}{T_A}\right)^2
=
2^3\times\frac14=2
\]
Hence,
\[
\frac{T_B}{T_A}=\sqrt2
\]
\(\textbf{Use angular momentum conservation within B's orbit:}\)
At B's perihelion and aphelion,
\[
r_pv_p=r_av_a
\]
Thus,
\[
\frac{v_p}{v_a}
=
\frac{14r}{2r}=7
\]
\(\textbf{Final answer:}\)
\[
\boxed{\frac{T_B}{T_A}=\sqrt2,\qquad
\frac{v_p}{v_a}=7}
\]
208. At point P of a planet's orbit, its distance from the Sun is \(2r\), its speed is \(3v\), and its velocity makes an angle \(30^\circ\) with the outward radius vector. At point Q of the same orbit, its speed is \(2v\) and its velocity is perpendicular to the radius vector. What are the Sun-planet distance \(r_Q\) and the ratio of instantaneous angular speeds \(\omega_Q/\omega_P\)?
ⓐ. \(r_Q=3r,\quad \omega_Q/\omega_P=4/9\)
ⓑ. \(r_Q=\dfrac{3r}{2},\quad \omega_Q/\omega_P=\dfrac{16}{9}\)
ⓒ. \(r_Q=\dfrac{3r}{2},\quad \omega_Q/\omega_P=\dfrac43\)
ⓓ. \(r_Q=2r,\quad \omega_Q/\omega_P=2\)
Correct Answer: \(r_Q=\dfrac{3r}{2},\quad \omega_Q/\omega_P=\dfrac{16}{9}\)
Explanation: \(\textbf{Extract the transverse speed at P:}\)
Only the velocity component perpendicular to the radius contributes to angular momentum.
Thus,
\[
v_{\perp P}=3v\sin30^\circ
\]
Hence,
\[
v_{\perp P}=\frac{3v}{2}
\]
At Q the velocity is perpendicular to the radius, so
\[
v_{\perp Q}=2v
\]
\(\textbf{Use conservation of angular momentum:}\)
\[
r_Pv_{\perp P}=r_Qv_{\perp Q}
\]
Therefore,
\[
(2r)\left(\frac{3v}{2}\right)
=
r_Q(2v)
\]
Thus,
\[
3rv=2vr_Q,
\]
giving
\[
r_Q=\frac{3r}{2}
\]
\(\textbf{Compare the angular speeds:}\)
Instantaneous angular speed is
\[
\omega=\frac{v_\perp}{r}
\]
Thus,
\[
\omega_P
=
\frac{3v/2}{2r}
=
\frac{3v}{4r},
\]
while
\[
\omega_Q
=
\frac{2v}{3r/2}
=
\frac{4v}{3r}
\]
Therefore,
\[
\frac{\omega_Q}{\omega_P}
=
\frac{4/3}{3/4}
=
\frac{16}{9}
\]
\(\textbf{Final answer:}\)
\[
\boxed{r_Q=\frac{3r}{2},\qquad
\frac{\omega_Q}{\omega_P}=\frac{16}{9}}
\]
209. A body of mass \(m\) moves in a circular orbit around a central mass \(M\). The magnitude of the gravitational force on the orbiting body is measured to be \(F\), and its orbital period is \(T\). Which pair correctly gives the orbital radius \(r\) and the central mass \(M\)?
ⓐ. \(r=\dfrac{FT^2}{4\pi^2m},\quad M=\dfrac{F^3T^4}{16\pi^4Gm^3}\)
ⓑ. \(r=\dfrac{4\pi^2m}{FT^2},\quad M=\dfrac{FT^2}{4\pi^2Gm}\)
ⓒ. \(r=\dfrac{FT}{2\pi m},\quad M=\dfrac{F^2T^2}{4\pi^2Gm^2}\)
ⓓ. \(r=\dfrac{FT^2}{2\pi m},\quad M=\dfrac{F^3T^4}{4\pi^2Gm^3}\)
Correct Answer: \(r=\dfrac{FT^2}{4\pi^2m},\quad M=\dfrac{F^3T^4}{16\pi^4Gm^3}\)
Explanation: \(\textbf{Use the measured force as the centripetal force:}\)
For circular motion,
\[
F=\frac{mv^2}{r}
\]
With
\[
v=\frac{2\pi r}{T},
\]
we obtain
\[
F
=
\frac{m(2\pi r/T)^2}{r}
\]
Therefore,
\[
F=\frac{4\pi^2mr}{T^2}
\]
Hence,
\[
r=\frac{FT^2}{4\pi^2m}
\]
\(\textbf{Recover the central mass:}\)
Newton's gravitational law gives
\[
F=\frac{GMm}{r^2}
\]
Thus,
\[
M=\frac{Fr^2}{Gm}
\]
Substituting the inferred radius,
\[
M
=
\frac{F}{Gm}
\left(\frac{FT^2}{4\pi^2m}\right)^2
\]
Therefore,
\[
M
=
\frac{F^3T^4}{16\pi^4Gm^3}
\]
Both expressions are based on the same gravitational force simultaneously providing the required centripetal force.
\(\textbf{Final answer:}\)
\[
\boxed{r=\frac{FT^2}{4\pi^2m},\qquad
M=\frac{F^3T^4}{16\pi^4Gm^3}}
\]
210. A planet has an elliptical orbit of semi-major axis \(a\) and period \(T\) around a star of true mass \(M\). An observer mistakenly uses the full major-axis length \(2a\) in place of the semi-major axis in
\[
M=\frac{4\pi^2a^3}{GT^2}.
\]
What mass will the observer infer?
ⓐ. \(M/8\)
ⓑ. \(2M\)
ⓒ. \(4M\)
ⓓ. \(8M\)
Correct Answer: \(8M\)
Explanation: \(\textbf{Correct central-mass relation:}\)
Using the proper semi-major axis,
\[
M=\frac{4\pi^2a^3}{GT^2}
\]
The observer incorrectly substitutes
\[
a_{\text{wrong}}=2a
\]
Thus the calculated mass is
\[
M_{\text{calc}}
=
\frac{4\pi^2(2a)^3}{GT^2}
\]
Since
\[
(2a)^3=8a^3,
\]
we obtain
\[
M_{\text{calc}}
=
8\frac{4\pi^2a^3}{GT^2}
\]
Therefore,
\[
M_{\text{calc}}=8M
\]
The error is cubed because the orbital-size quantity enters Kepler's third law through \(a^3\).
The full major axis and the semi-major axis must therefore not be interchanged.
\(\textbf{Final answer:}\)
\[
\boxed{M_{\text{calc}}=8M}
\]
211. A body of mass \(m\) moves in a circular orbit of radius \(r\) around a star and experiences gravitational force \(F\). It is replaced by another body of mass \(5m\), which is given the correct speed for the same circular orbit. Neglect both orbiting masses compared with the star. Which set of changes is correct?
ⓐ. Gravitational force becomes \(5F\), orbital acceleration remains unchanged, and orbital period remains unchanged
ⓑ. Gravitational force remains \(F\), orbital acceleration becomes five times larger, and orbital period remains unchanged
ⓒ. Gravitational force becomes \(5F\), orbital acceleration becomes five times larger, and orbital period becomes five times larger
ⓓ. Gravitational force becomes \(F/5\), orbital acceleration remains unchanged, and orbital period becomes \(T/\sqrt5\)
Correct Answer: Gravitational force becomes \(5F\), orbital acceleration remains unchanged, and orbital period remains unchanged
Explanation: For the original body,
\[
F=\frac{GMm}{r^2}
\]
Replacing \(m\) by \(5m\) at the same orbital radius gives
\[
F'=\frac{GM(5m)}{r^2}=5F
\]
Thus the gravitational force does increase with orbiting-body mass.
However, the resulting acceleration is
\[
a=\frac{F}{m}=\frac{GM}{r^2}
\]
For the replacement body,
\[
a'=\frac{5F}{5m}=a
\]
Therefore the required circular acceleration is unchanged.
The circular orbital speed satisfies
\[
v=\sqrt{\frac{GM}{r}},
\]
which contains no orbiting-body mass.
Hence the correctly established replacement orbit has the same speed.
Its period
\[
T=\frac{2\pi r}{v}
\]
is consequently unchanged as well.
The orbiting mass cancels from the ideal circular-orbit dynamics even though the gravitational force itself is proportional to that mass.
Therefore option A is correct.
212. Planet P moves around a star in an elliptical orbit of semi-major axis \(4r\). Its perihelion speed is three times its aphelion speed. Planet Q revolves around the same star and has an orbital period eight times that of P. What are the perihelion and aphelion distances of P, and the semi-major axis of Q?
ⓐ. \(r_p=2r,\ r_a=6r,\ a_Q=16r\)
ⓑ. \(r_p=r,\ r_a=7r,\ a_Q=32r\)
ⓒ. \(r_p=2r,\ r_a=6r,\ a_Q=8r\)
ⓓ. \(r_p=3r,\ r_a=5r,\ a_Q=16r\)
Correct Answer: \(r_p=2r,\ r_a=6r,\ a_Q=16r\)
Explanation: \(\textbf{Use the speed ratio to reconstruct P's ellipse:}\)
At perihelion and aphelion, conservation of angular momentum gives
\[
r_pv_p=r_av_a
\]
Therefore,
\[
\frac{r_a}{r_p}=\frac{v_p}{v_a}=3
\]
Hence,
\[
r_a=3r_p
\]
The semi-major axis of P is \(4r\), so
\[
r_p+r_a=2a_P=8r
\]
Substituting \(r_a=3r_p\),
\[
4r_p=8r
\]
Thus,
\[
r_p=2r,\qquad r_a=6r
\]
\(\textbf{Use the period change to determine Q's orbital size:}\)
For planets orbiting the same star,
\[
\frac{T_Q}{T_P}
=
\left(\frac{a_Q}{a_P}\right)^{3/2}
\]
Given
\[
\frac{T_Q}{T_P}=8,
\]
we obtain
\[
8=\left(\frac{a_Q}{4r}\right)^{3/2}
\]
Raising both sides to the power \(2/3\),
\[
\frac{a_Q}{4r}=4
\]
Hence,
\[
a_Q=16r
\]
\(\textbf{Final answer:}\)
\[
\boxed{r_p=2r,\qquad r_a=6r,\qquad a_Q=16r}
\]
213. A small body moves in a circular orbit just above the surface of a uniform spherical planet. Its orbital period is \(T\), and the planet's surface acceleration due to gravity is \(g\). Which pair correctly gives the planetary radius \(R\) and mean density \(\rho\)?
ⓐ. \(R=\dfrac{gT^2}{2\pi},\quad \rho=\dfrac{3\pi T^2}{G}\)
ⓑ. \(R=\dfrac{4\pi^2g}{T^2},\quad \rho=\dfrac{3}{4\pi GT^2}\)
ⓒ. \(R=\dfrac{gT^2}{4\pi^2},\quad \rho=\dfrac{3\pi}{GT^2}\)
ⓓ. \(R=\dfrac{gT^2}{4\pi^2},\quad \rho=\dfrac{GT^2}{3\pi}\)
Correct Answer: \(R=\dfrac{gT^2}{4\pi^2},\quad \rho=\dfrac{3\pi}{GT^2}\)
Explanation: \(\textbf{Relate surface gravity to circular motion:}\)
Just above the surface, the orbital radius is \(R\).
Gravity supplies the centripetal acceleration:
\[
g=\frac{v^2}{R}
\]
For the circular orbit,
\[
v=\frac{2\pi R}{T}
\]
Therefore,
\[
g
=
\frac{(2\pi R/T)^2}{R}
\]
Hence,
\[
g=\frac{4\pi^2R}{T^2}
\]
Solving for \(R\),
\[
R=\frac{gT^2}{4\pi^2}
\]
\(\textbf{Eliminate radius to determine the density:}\)
For a uniform sphere,
\[
M=\frac43\pi R^3\rho
\]
The circular-period relation is
\[
T^2=\frac{4\pi^2R^3}{GM}
\]
Substituting the mass expression,
\[
T^2
=
\frac{4\pi^2R^3}
{G(4\pi R^3\rho/3)}
\]
Thus,
\[
T^2=\frac{3\pi}{G\rho}
\]
Therefore,
\[
\rho=\frac{3\pi}{GT^2}
\]
\(\textbf{Final answer:}\)
\[
\boxed{R=\frac{gT^2}{4\pi^2},\qquad
\rho=\frac{3\pi}{GT^2}}
\]
214. Uniform spherical planets A and B have the same surface acceleration due to gravity, but \(R_B=4R_A\). Small bodies are placed in circular orbits just above their respective surfaces. If the orbital periods are \(T_A,T_B\), what are \(T_B/T_A\) and \(M_B/M_A\)?
ⓐ. \(2,\ 16\)
ⓑ. \(4,\ 16\)
ⓒ. \(2,\ 4\)
ⓓ. \(4,\ 64\)
Correct Answer: \(2,\ 16\)
Explanation: \(\textbf{Compare the near-surface orbital periods:}\)
For a circular orbit just above a planet's surface,
\[
T=2\pi\sqrt{\frac{R^3}{GM}}
\]
Surface gravity satisfies
\[
g=\frac{GM}{R^2}
\]
Thus,
\[
GM=gR^2
\]
Substituting into the period relation,
\[
T=2\pi\sqrt{\frac{R^3}{gR^2}}
\]
Therefore,
\[
T=2\pi\sqrt{\frac{R}{g}}
\]
The two planets have the same \(g\), so
\[
T\propto\sqrt R
\]
Hence,
\[
\frac{T_B}{T_A}
=
\sqrt{\frac{R_B}{R_A}}
=\sqrt4=2
\]
\(\textbf{Compare the planetary masses:}\)
At equal surface \(g\),
\[
M=\frac{gR^2}{G}
\]
Therefore,
\[
M\propto R^2
\]
Thus,
\[
\frac{M_B}{M_A}
=
\left(\frac{R_B}{R_A}\right)^2
=4^2=16
\]
\(\textbf{Final answer:}\)
\[
\boxed{\frac{T_B}{T_A}=2,\qquad
\frac{M_B}{M_A}=16}
\]
215. Assertion: For any two arbitrary points P and Q of the same elliptical planetary orbit, the relation \(r_Pv_P=r_Qv_Q\) must hold, where \(v\) denotes the total speed.
Reason: Conservation of angular momentum under the central gravitational force gives \(rv_\perp=\text{constant}\), where \(v_\perp\) is the component of velocity perpendicular to the Sun-planet radius.
ⓐ. Both Assertion and Reason are correct, and Reason correctly explains Assertion
ⓑ. Both Assertion and Reason are correct, but Reason does not correctly explain Assertion
ⓒ. Assertion is correct but Reason is incorrect
ⓓ. Assertion is incorrect but Reason is correct
Correct Answer: Assertion is incorrect but Reason is correct
Explanation: The gravitational force is directed along the Sun-planet radius and is therefore a central force.
Its torque about the Sun is zero.
Consequently, angular momentum about the Sun is conserved.
For a planet of mass \(m\),
\[
L=mr v_\perp.
\]
Therefore,
\[
rv_\perp=\text{constant}.
\]
At a general point of an elliptical orbit, the velocity need not be perpendicular to the radius vector.
Hence the total speed \(v\) cannot generally replace \(v_\perp\).
The special relation
\[
r_pv_p=r_av_a
\]
is valid at perihelion and aphelion because the velocity there is transverse to the radius.
Thus the Reason is correct, while the Assertion overextends an apsidal relation to arbitrary orbital points.
216. Two small bodies move in different circular orbits around the same star. Their orbital speeds satisfy \(v_1=2v_2\). What are \(r_1/r_2\) and \(T_1/T_2\)?
ⓐ. \(\dfrac14,\ \dfrac18\)
ⓑ. \(\dfrac12,\ \dfrac14\)
ⓒ. \(\dfrac14,\ \dfrac14\)
ⓓ. \(4,\ 8\)
Correct Answer: \(\dfrac14,\ \dfrac18\)
Explanation: \(\textbf{Use the circular-speed dependence:}\)
For circular gravitational motion around the same star,
\[
v=\sqrt{\frac{GM}{r}}
\]
Therefore,
\[
v\propto r^{-1/2}
\]
Hence,
\[
\frac{v_1}{v_2}
=
\sqrt{\frac{r_2}{r_1}}
\]
Given
\[
\frac{v_1}{v_2}=2,
\]
we obtain
\[
4=\frac{r_2}{r_1}
\]
Thus,
\[
\frac{r_1}{r_2}=\frac14
\]
\(\textbf{Now compare the periods:}\)
Kepler's third law gives
\[
T\propto r^{3/2}
\]
Therefore,
\[
\frac{T_1}{T_2}
=
\left(\frac{r_1}{r_2}\right)^{3/2}
\]
Substituting,
\[
\frac{T_1}{T_2}
=
\left(\frac14\right)^{3/2}
=\frac18
\]
Equivalently, the circular-orbit invariant
\[
v^3T=2\pi GM
\]
also gives \(T_1/T_2=1/8\).
\(\textbf{Final answer:}\)
\[
\boxed{\frac{r_1}{r_2}=\frac14,\qquad
\frac{T_1}{T_2}=\frac18}
\]
217. A small body is claimed to move in a circular orbit in each of the following observations. Take \(4\pi^2\approx40\).
| Observation | Orbital radius \(r\) | Period \(T\) | Reported gravitational acceleration \(g_o\) |
|---|
| P | \(1.0\times10^7\,\text{m}\) | \(1.0\times10^4\,\text{s}\) | \(4.0\,\text{m s}^{-2}\) |
| Q | \(2.0\times10^7\,\text{m}\) | \(2.0\times10^4\,\text{s}\) | \(2.0\,\text{m s}^{-2}\) |
| R | \(4.0\times10^7\,\text{m}\) | \(4.0\times10^4\,\text{s}\) | \(1.0\,\text{m s}^{-2}\) |
| S | \(9.0\times10^7\,\text{m}\) | \(3.0\times10^4\,\text{s}\) | \(3.0\,\text{m s}^{-2}\) |
Which observation is inconsistent with circular motion, and what should its reported acceleration be?
ⓐ. P; \(2.0\,\text{m s}^{-2}\)
ⓑ. Q; \(4.0\,\text{m s}^{-2}\)
ⓒ. R; \(2.0\,\text{m s}^{-2}\)
ⓓ. S; \(4.0\,\text{m s}^{-2}\)
Correct Answer: S; \(4.0\,\text{m s}^{-2}\)
Explanation: \(\textbf{Construct the compatibility relation:}\)
For circular motion,
\[
g_o=\frac{v^2}{r}.
\]
With
\[
v=\frac{2\pi r}{T},
\]
we obtain
\[
g_o=\frac{4\pi^2r}{T^2}.
\]
Using \(4\pi^2\approx40\), check each row.
For P,
\[
g_o
=
\frac{40(1.0\times10^7)}
{(1.0\times10^4)^2}
=4.0\,\text{m s}^{-2}.
\]
For Q,
\[
g_o
=
\frac{40(2.0\times10^7)}
{(2.0\times10^4)^2}
=2.0\,\text{m s}^{-2}.
\]
For R,
\[
g_o
=
\frac{40(4.0\times10^7)}
{(4.0\times10^4)^2}
=1.0\,\text{m s}^{-2}.
\]
For S,
\[
g_o
=
\frac{40(9.0\times10^7)}
{(3.0\times10^4)^2}.
\]
Hence,
\[
g_o=4.0\,\text{m s}^{-2}.
\]
Thus S is the inconsistent observation.
\(\textbf{Final answer:}\) S should report \(4.0\,\text{m s}^{-2}\).
218. At two points P and Q of the same planetary orbit, the instantaneous angular speeds about the Sun are \(9\omega\) and \(4\omega\), respectively. What are \(r_Q/r_P\) and \(v_{\perp Q}/v_{\perp P}\)?
ⓐ. \(\dfrac32,\ \dfrac32\)
ⓑ. \(\dfrac94,\ \dfrac49\)
ⓒ. \(\dfrac32,\ \dfrac23\)
ⓓ. \(\dfrac23,\ \dfrac32\)
Correct Answer: \(\dfrac32,\ \dfrac23\)
Explanation: \(\textbf{Use constant areal rate in angular form:}\)
Kepler's second law gives
\[
\frac{dA}{dt}
=
\frac12r^2\omega
=
\text{constant}
\]
Therefore,
\[
r_P^2\omega_P=r_Q^2\omega_Q
\]
Using
\[
\omega_P=9\omega,
\qquad
\omega_Q=4\omega,
\]
we have
\[
9r_P^2=4r_Q^2
\]
Thus,
\[
\frac{r_Q^2}{r_P^2}=\frac94
\]
Hence,
\[
\frac{r_Q}{r_P}=\frac32
\]
\(\textbf{Convert to transverse speeds:}\)
The transverse component satisfies
\[
v_\perp=r\omega
\]
Therefore,
\[
\frac{v_{\perp Q}}{v_{\perp P}}
=
\frac{r_Q\omega_Q}{r_P\omega_P}
\]
Substitution gives
\[
\frac{v_{\perp Q}}{v_{\perp P}}
=
\frac{(3/2)(4)}{9}
=\frac23
\]
\(\textbf{Final answer:}\)
\[
\boxed{\frac{r_Q}{r_P}=\frac32,\qquad
\frac{v_{\perp Q}}{v_{\perp P}}=\frac23}
\]
219. Small bodies A and B move in circular orbits in two different gravitational systems. At B's orbit, the gravitational acceleration is half that at A's orbit, while \(T_B=2T_A\). What are \(r_B/r_A\) and \(M_B/M_A\)?
ⓐ. \(1/2,\ 2\)
ⓑ. \(2,\ 1/2\)
ⓒ. \(4,\ 2\)
ⓓ. \(2,\ 2\)
Correct Answer: \(2,\ 2\)
Explanation: \(\textbf{Infer the orbital-radius ratio from acceleration and period:}\)
For circular motion,
\[
g_o=\frac{4\pi^2r}{T^2}
\]
Therefore,
\[
r=\frac{g_oT^2}{4\pi^2}
\]
Hence,
\[
\frac{r_B}{r_A}
=
\frac{g_B}{g_A}
\left(\frac{T_B}{T_A}\right)^2
\]
Given
\[
\frac{g_B}{g_A}=\frac12,
\qquad
\frac{T_B}{T_A}=2,
\]
we obtain
\[
\frac{r_B}{r_A}
=
\frac12(2)^2=2
\]
\(\textbf{Use the local gravitational acceleration to compare central masses:}\)
At the orbital radius,
\[
g_o=\frac{GM}{r^2}
\]
Thus,
\[
M=\frac{g_or^2}{G}
\]
Therefore,
\[
\frac{M_B}{M_A}
=
\frac{g_B}{g_A}
\left(\frac{r_B}{r_A}\right)^2
\]
Hence,
\[
\frac{M_B}{M_A}
=
\frac12(2)^2=2
\]
\(\textbf{Final answer:}\)
\[
\boxed{\frac{r_B}{r_A}=2,\qquad
\frac{M_B}{M_A}=2}
\]
220. Two geometrically identical elliptical orbits are considered around stars A and B of masses \(M\) and \(9M\), respectively. A planet of negligible mass occupies each orbit. What are \(T_B/T_A\), the ratio of their areal rates \(\dot A_B/\dot A_A\), and the ratio of their transverse speeds at corresponding points?
ⓐ. \(\dfrac13,\ 3,\ 3\)
ⓑ. \(3,\ \dfrac13,\ \dfrac13\)
ⓒ. \(\dfrac13,\ 9,\ 3\)
ⓓ. \(\dfrac19,\ 3,\ 9\)
Correct Answer: \(\dfrac13,\ 3,\ 3\)
Explanation: \(\textbf{Compare orbital periods for identical semi-major axes:}\)
The two ellipses are geometrically identical, so
\[
a_B=a_A
\]
Newton's form of Kepler's third law gives
\[
T^2\propto\frac{a^3}{M}
\]
Thus,
\[
\frac{T_B}{T_A}
=
\sqrt{\frac{M_A}{M_B}}
\]
With
\[
M_B=9M_A,
\]
\[
\frac{T_B}{T_A}=\frac13
\]
\(\textbf{Compare the areal rates:}\)
The two geometric ellipses have the same total area.
Each planet sweeps the entire orbital area once per period.
Therefore,
\[
\dot A\propto\frac1T
\]
Hence,
\[
\frac{\dot A_B}{\dot A_A}=3
\]
\(\textbf{Compare corresponding transverse speeds:}\)
At corresponding points the radii are equal.
Since
\[
\dot A=\frac12rv_\perp,
\]
the threefold areal-rate ratio implies
\[
\frac{v_{\perp B}}{v_{\perp A}}=3
\]
\(\textbf{Final answer:}\)
\[
\boxed{\frac{T_B}{T_A}=\frac13,\qquad
\frac{\dot A_B}{\dot A_A}=3,\qquad
\frac{v_{\perp B}}{v_{\perp A}}=3}
\]
