501. A uniform spherical planet has radius \(R\) and surface gravitational acceleration \(g\). Astronaut P stands on a supported platform at depth \(d\) below the surface. Astronaut Q of the same mass is freely floating in a circular spacecraft at centre-distance \(2R\). The actual gravitational forces on P and Q are equal. Neglect tidal effects. Which statement is correct?
ⓐ. \(d=R/4\), and both scale readings are \(mg/4\)
ⓑ. \(d=3R/4\), P's scale reads \(mg/4\), while Q's scale reads zero
ⓒ. \(d=3R/4\), and both scale readings are zero
ⓓ. \(d=R/2\), P's scale reads \(mg/2\), while Q's scale reads \(mg/4\)
Correct Answer: \(d=3R/4\), P's scale reads \(mg/4\), while Q's scale reads zero
Explanation: \(\textbf{Find the gravitational acceleration at the circular orbit:}\)
At centre-distance \(2R\),
\[
g_Q=g\left(\frac{R}{2R}\right)^2
\]
Hence,
\[
g_Q=\frac g4
\]
\(\textbf{Match this with gravity at depth:}\)
For a uniform spherical planet,
\[
g_d=g\left(1-\frac dR\right)
\]
Equal gravitational forces for equal astronaut masses require
\[
g_d=g_Q
\]
Thus,
\[
1-\frac dR=\frac14
\]
Therefore,
\[
d=\frac{3R}{4}
\]
\(\textbf{Distinguish actual gravitational force from apparent weight:}\)
Astronaut P is supported, so the platform must provide
\[
N_P=mg_d=\frac{mg}{4}
\]
Astronaut Q and the spacecraft freely fall together.
Therefore,
\[
N_Q=0
\]
The actual gravitational force on Q is still
\[
\frac{mg}{4},
\]
but the apparent weight is zero.
\(\textbf{Final answer:}\)
\[
d=\frac{3R}{4},\qquad
N_P=\frac{mg}{4},\qquad
N_Q=0
\]
502. Uniform spherical planets A and B have equal surface gravitational potentials, while
\[
\rho_B=4\rho_A.
\]
For satellites moving just above their respective surfaces, which set is correct?
ⓐ. \(R_B/R_A=1/2,\quad M_B/M_A=2,\quad g_B/g_A=1,\quad T_B/T_A=2\)
ⓑ. \(R_B/R_A=1/4,\quad M_B/M_A=1/4,\quad g_B/g_A=4,\quad T_B/T_A=1/4\)
ⓒ. \(R_B/R_A=2,\quad M_B/M_A=8,\quad g_B/g_A=2,\quad T_B/T_A=2\)
ⓓ. \(R_B/R_A=1/2,\quad M_B/M_A=1/2,\quad g_B/g_A=2,\quad T_B/T_A=1/2\)
Correct Answer: \(R_B/R_A=1/2,\quad M_B/M_A=1/2,\quad g_B/g_A=2,\quad T_B/T_A=1/2\)
Explanation: \(\textbf{Use surface potential in terms of density and radius:}\)
For a uniform sphere,
\[
M=\frac43\pi\rho R^3
\]
The surface potential magnitude is
\[
|V_s|=\frac{GM}{R}
\]
Therefore,
\[
|V_s|
=
\frac43\pi G\rho R^2
\]
Equal surface potentials imply
\[
\rho_AR_A^2=\rho_BR_B^2
\]
Given
\[
\rho_B=4\rho_A,
\]
\[
R_A^2=4R_B^2
\]
Thus,
\[
\frac{R_B}{R_A}=\frac12
\]
\(\textbf{Compare masses:}\)
\[
M\propto\rho R^3
\]
Hence,
\[
\frac{M_B}{M_A}
=
4\left(\frac12\right)^3
=
\frac12
\]
\(\textbf{Compare surface gravities:}\)
\[
g\propto\rho R
\]
Therefore,
\[
\frac{g_B}{g_A}
=
4\left(\frac12\right)=2
\]
\(\textbf{Compare near-surface orbital periods:}\)
\[
T^2=\frac{3\pi}{G\rho}
\]
Hence,
\[
\frac{T_B}{T_A}
=
\sqrt{\frac{\rho_A}{\rho_B}}
=
\frac12
\]
\(\textbf{Final answer:}\)
\[
\boxed{\frac{R_B}{R_A}=\frac12,\quad
\frac{M_B}{M_A}=\frac12,\quad
\frac{g_B}{g_A}=2,\quad
\frac{T_B}{T_A}=\frac12}
\]
503. A body of mass \(m\) is raised quasistatically from the surface of a spherical planet to altitude \(h\). At the final radius, the work already done in raising it is equal to the binding energy that the same body would have if placed in a circular orbit there. If \(v_e\) is the escape speed from the planet's surface, what are \(h\) and the circular orbital speed at the final radius?
ⓐ. \(h=R/2,\quad v_o=v_e/\sqrt3\)
ⓑ. \(h=R,\quad v_o=v_e/2\)
ⓒ. \(h=2R,\quad v_o=v_e/\sqrt6\)
ⓓ. \(h=R/2,\quad v_o=v_e/\sqrt2\)
Correct Answer: \(h=R/2,\quad v_o=v_e/\sqrt3\)
Explanation: \(\textbf{Write the quasistatic raising work:}\)
Let the final centre-distance be
\[
r=R+h
\]
The work required to raise the body from \(R\) to \(r\) is
\[
W
=
GMm\left(\frac1R-\frac1r\right)
\]
\(\textbf{Write the circular-orbit binding energy at the same radius:}\)
For a circular satellite,
\[
B=\frac{GMm}{2r}
\]
The question gives
\[
W=B
\]
Therefore,
\[
\frac1R-\frac1r=\frac1{2r}
\]
Hence,
\[
\frac1R=\frac3{2r}
\]
Thus,
\[
r=\frac{3R}{2}
\]
Therefore,
\[
h=r-R=\frac R2
\]
\(\textbf{Compare final orbital speed with surface escape speed:}\)
At the final radius,
\[
v_o^2=\frac{GM}{3R/2}
=\frac{2GM}{3R}
\]
At the surface,
\[
v_e^2=\frac{2GM}{R}
\]
Hence,
\[
\frac{v_o^2}{v_e^2}=\frac13
\]
Therefore,
\[
v_o=\frac{v_e}{\sqrt3}
\]
\(\textbf{Final answer:}\)
\[
\boxed{h=\frac R2,\qquad
v_o=\frac{v_e}{\sqrt3}}
\]
504. For circular satellites around the same central body, let \(v_e\) denote the escape speed from each satellite's orbital radius. A graph is constructed as follows.
The vertical axis is \(Y=v_e^3\), and the horizontal axis is \(X=1/T\), where \(T\) is the circular orbital period at that same radius. The graph is a straight line through the origin with slope \(S\).
Which pair is correct?
ⓐ. \(M=\dfrac{4\sqrt2\pi S}{G},\quad v_e=(x_0/S)^{1/3}\)
ⓑ. \(M=\dfrac{S}{2\sqrt2\pi G},\quad v_e=(Sx_0)^{1/2}\)
ⓒ. \(M=\dfrac{S}{8\sqrt2\pi G},\quad v_e=Sx_0\)
ⓓ. \(M=\dfrac{S}{4\sqrt2\pi G},\quad v_e=(Sx_0)^{1/3}\)
Correct Answer: \(M=\dfrac{S}{4\sqrt2\pi G},\quad v_e=(Sx_0)^{1/3}\)
Explanation: \(\textbf{Connect local escape speed with circular orbital speed:}\)
At the same radius,
\[
v_e=\sqrt2\,v_o
\]
For circular satellites,
\[
v_o^3T=2\pi GM
\]
Therefore,
\[
v_e^3T
=
(\sqrt2)^3v_o^3T
\]
Since
\[
(\sqrt2)^3=2\sqrt2,
\]
\[
v_e^3T
=
2\sqrt2(2\pi GM)
\]
Hence,
\[
v_e^3T=4\sqrt2\pi GM
\]
Thus,
\[
v_e^3
=
4\sqrt2\pi GM\left(\frac1T\right)
\]
The graph slope is therefore
\[
S=4\sqrt2\pi GM
\]
Hence,
\[
M=\frac{S}{4\sqrt2\pi G}
\]
At
\[
X=x_0,
\]
\[
v_e^3=Sx_0
\]
Therefore,
\[
v_e=(Sx_0)^{1/3}
\]
\(\textbf{Final answer:}\)
\[
\boxed{M=\frac{S}{4\sqrt2\pi G},\qquad
v_e=(Sx_0)^{1/3}}
\]
505. Precise observations of a circular satellite determine the gravitational parameter
\[
\mu=GM
\]
of a planet. A laboratory value of \(G\) used by a student is \(4\%\) larger than the true value, while the satellite data and planetary radius are exact. Which statement is correct?
ⓐ. \(M_{\text{inf}}=\dfrac{25}{26}M_{\text{true}},\quad g_{\text{calc}}=g_{\text{true}}\)
ⓑ. \(M_{\text{inf}}=\dfrac{26}{25}M_{\text{true}},\quad g_{\text{calc}}=\dfrac{26}{25}g_{\text{true}}\)
ⓒ. \(M_{\text{inf}}=M_{\text{true}},\quad g_{\text{calc}}=\dfrac{25}{26}g_{\text{true}}\)
ⓓ. \(M_{\text{inf}}=\dfrac{25}{26}M_{\text{true}},\quad g_{\text{calc}}=\dfrac{25}{26}g_{\text{true}}\)
Correct Answer: \(M_{\text{inf}}=\dfrac{25}{26}M_{\text{true}},\quad g_{\text{calc}}=g_{\text{true}}\)
Explanation: \(\textbf{Separate the directly observed gravitational parameter from }G\textbf{:}\)
Satellite observations can determine
\[
\mu=GM
\]
without first separating \(G\) and \(M\).
Let the true value of the universal constant be \(G\).
The student uses
\[
G_{\text{used}}=1.04G=\frac{26}{25}G
\]
The inferred mass is
\[
M_{\text{inf}}
=
\frac{\mu}{G_{\text{used}}}
\]
Since
\[
\mu=GM_{\text{true}},
\]
\[
M_{\text{inf}}
=
\frac{GM_{\text{true}}}{(26/25)G}
\]
Therefore,
\[
M_{\text{inf}}
=
\frac{25}{26}M_{\text{true}}
\]
\(\textbf{Check the surface-gravity prediction:}\)
If the planet's radius is known,
\[
g_s=\frac{GM}{R^2}
=\frac{\mu}{R^2}
\]
Both \(\mu\) and \(R\) are exact in the question.
Therefore the predicted surface gravity is correct despite the incorrect separated value assigned to \(M\).
\(\textbf{Final answer:}\)
\[
M_{\text{inf}}=\frac{25}{26}M_{\text{true}},
\]
while \(g_s\) obtained from \(\mu/R^2\) is unchanged.
506. A satellite of mass \(m\) moves in a circular orbit with binding energy \(B\) and period \(T\). Which pair correctly gives the magnitude of the gravitational force \(F\) on the satellite and the central gravitational parameter \(GM\)?
ⓐ. \(F=\dfrac{2\sqrt2\pi\sqrt{mB}}{T},\quad GM=\dfrac{T}{2\pi}\left(\dfrac{2B}{m}\right)^{3/2}\)
ⓑ. \(F=\dfrac{2\pi B}{T},\quad GM=\dfrac{BT^2}{m}\)
ⓒ. \(F=\dfrac{\sqrt2\pi\sqrt{mB}}{T},\quad GM=\dfrac{T}{\pi}\left(\dfrac{B}{m}\right)^{3/2}\)
ⓓ. \(F=\dfrac{4\pi mB}{T^2},\quad GM=\dfrac{2BT}{\pi m}\)
Correct Answer: \(F=\dfrac{2\sqrt2\pi\sqrt{mB}}{T},\quad GM=\dfrac{T}{2\pi}\left(\dfrac{2B}{m}\right)^{3/2}\)
Explanation: \(\textbf{Recover orbital speed from binding energy:}\)
For a circular satellite,
\[
B=K=\frac12mv^2
\]
Therefore,
\[
v=\sqrt{\frac{2B}{m}}
\]
\(\textbf{Use the period to infer orbital radius:}\)
\[
T=\frac{2\pi r}{v}
\]
Hence,
\[
r=\frac{vT}{2\pi}
\]
\(\textbf{Calculate the gravitational force:}\)
For circular motion,
\[
F=\frac{mv^2}{r}
\]
Substituting
\[
mv^2=2B
\]
and
\[
r=\frac{vT}{2\pi},
\]
\[
F
=
\frac{2B}{vT/(2\pi)}
\]
Thus,
\[
F=\frac{4\pi B}{vT}
\]
Using
\[
v=\sqrt{\frac{2B}{m}},
\]
\[
F=\frac{2\sqrt2\pi\sqrt{mB}}{T}
\]
\(\textbf{Recover the central gravitational parameter:}\)
For a circular orbit,
\[
GM=v^2r
\]
Therefore,
\[
GM
=
v^2\frac{vT}{2\pi}
=
\frac{v^3T}{2\pi}
\]
Hence,
\[
GM
=
\frac{T}{2\pi}
\left(\frac{2B}{m}\right)^{3/2}
\]
\(\textbf{Final answer:}\)
\[
\boxed{F=\frac{2\sqrt2\pi\sqrt{mB}}{T},\qquad
GM=\frac{T}{2\pi}\left(\frac{2B}{m}\right)^{3/2}}
\]
507. A satellite of mass \(m\) moves initially in a circular orbit of centre-distance \(2R\) around a spherical planet of radius \(R\) and surface gravity \(g\). An external mechanism finally brings it to rest on the planetary surface. Neglect dissipative losses other than energy deliberately removed by the mechanism. How much mechanical energy must be removed?
ⓐ. \(\dfrac14mgR\)
ⓑ. \(\dfrac12mgR\)
ⓒ. \(\dfrac34mgR\)
ⓓ. \(mgR\)
Correct Answer: \(\dfrac34mgR\)
Explanation: \(\textbf{Initial circular-orbit energy:}\)
At
\[
r=2R,
\]
the satellite total energy is
\[
E_i=-\frac{GMm}{2r}
\]
Therefore,
\[
E_i=-\frac{GMm}{4R}
\]
Using
\[
GM=gR^2,
\]
\[
E_i=-\frac14mgR
\]
\(\textbf{Final energy at rest on the surface:}\)
The final kinetic energy is zero.
Thus,
\[
E_f=-\frac{GMm}{R}
\]
Hence,
\[
E_f=-mgR
\]
The mechanical-energy change is
\[
\Delta E=E_f-E_i
\]
Therefore,
\[
\Delta E
=
-mgR+\frac14mgR
\]
Thus,
\[
\Delta E=-\frac34mgR
\]
The negative sign means energy must be removed.
Its required magnitude is
\[
\frac34mgR
\]
The initial circular-orbit binding energy is only \(mgR/4\), so the energy removed in reaching rest on the surface is three times that initial binding energy.
\(\textbf{Final answer:}\)
\[
\boxed{E_{\text{removed}}=\frac34mgR}
\]
508. A satellite of mass \(m\) initially moves in a circular orbit of radius \(r_0\) around mass \(M\). Let \(\Delta E\) be the net energy supplied to place it finally in another circular orbit of radius \(r\ge r_0\).
A graph is drawn with \(\Delta E\) on the vertical axis and \(x=1/r\) on the horizontal axis. The initial radius \(r_0\) is fixed.
Which description is correct?
ⓐ. Horizontal line; \(\Delta E\) is independent of the final radius
ⓑ. Straight line; slope \(+GMm/2\); vertical intercept \(0\)
ⓒ. Parabola; maximum value \(GMm/(2r_0)\) at \(x=0\)
ⓓ. Straight line; slope \(-GMm/2\); vertical intercept \(GMm/(2r_0)\)
Correct Answer: Straight line; slope \(-GMm/2\); vertical intercept \(GMm/(2r_0)\)
Explanation: \(\textbf{Write the two circular-orbit energies:}\)
Initially,
\[
E_i=-\frac{GMm}{2r_0}
\]
Finally,
\[
E_f=-\frac{GMm}{2r}
\]
Therefore,
\[
\Delta E=E_f-E_i
\]
Hence,
\[
\Delta E
=
-\frac{GMm}{2r}
+\frac{GMm}{2r_0}
\]
With
\[
x=\frac1r,
\]
this becomes
\[
\Delta E
=
-\frac{GMm}{2}x
+
\frac{GMm}{2r_0}
\]
Thus the graph is a straight line with slope
\[
-\frac{GMm}{2}
\]
At
\[
x=0,
\]
the final radius tends to infinity.
The graph intercept is then
\[
\frac{GMm}{2r_0},
\]
which is exactly the initial circular satellite's binding energy.
At
\[
x=\frac1{r_0},
\]
\[
\Delta E=0,
\]
as expected when the final and initial orbits are identical.
\(\textbf{Final answer:}\) The graph has slope \(-GMm/2\) and vertical intercept \(GMm/(2r_0)\).
509. A uniform spherical planet has radius \(R\) and surface gravitational acceleration \(g\). At a nonzero distance \(x\), the gravitational acceleration at altitude \(x\) above the surface is equal in magnitude to the gravitational acceleration at depth \(x\) below the surface. A satellite is placed in a circular orbit at that altitude. Which pair is correct?
ⓐ. \(x=R/2,\quad v_o=\sqrt{2gR/3}\)
ⓑ. \(x=\dfrac{\sqrt5+1}{2}R,\quad v_o=\sqrt{\dfrac{\sqrt5+1}{2}gR}\)
ⓒ. \(x=\dfrac{\sqrt5-1}{2}R,\quad v_o=\sqrt{\dfrac{\sqrt5-1}{2}gR}\)
ⓓ. \(x=(\sqrt2-1)R,\quad v_o=\sqrt{gR/2}\)
Correct Answer: \(x=\dfrac{\sqrt5-1}{2}R,\quad v_o=\sqrt{\dfrac{\sqrt5-1}{2}gR}\)
Explanation: \(\textbf{Set the altitude and depth expressions equal:}\)
Let
\[
y=\frac{x}{R}
\]
At altitude \(x\),
\[
\frac{g_h}{g}
=
\frac1{(1+y)^2}
\]
At depth \(x\) in a uniform planet,
\[
\frac{g_d}{g}=1-y
\]
The given condition is
\[
1-y=\frac1{(1+y)^2}
\]
Multiplying,
\[
(1-y)(1+y)^2=1
\]
Expanding and simplifying,
\[
y-y^2-y^3=0
\]
The nonzero solution satisfies
\[
y^2+y-1=0
\]
Hence,
\[
y=\frac{\sqrt5-1}{2}
\]
Therefore,
\[
x=\frac{\sqrt5-1}{2}R
\]
\(\textbf{Find the orbital speed at that altitude:}\)
The orbital centre-distance is
\[
r=R(1+y)
\]
Using
\[
GM=gR^2,
\]
\[
v_o^2=\frac{GM}{r}
=\frac{gR}{1+y}
\]
From
\[
y^2+y=1,
\]
we have
\[
\frac1{1+y}=y
\]
Thus,
\[
v_o^2=ygR
\]
Therefore,
\[
v_o=\sqrt{\frac{\sqrt5-1}{2}gR}
\]
\(\textbf{Final answer:}\)
\[
\boxed{x=\frac{\sqrt5-1}{2}R,\qquad
v_o=\sqrt{\frac{\sqrt5-1}{2}gR}}
\]
510. A satellite moves in a circular orbit at altitude \(2R\) above a spherical planet of radius \(R\). Its binding energy per unit mass is \(b\). Let \(T_0\) be the period of a circular satellite just above the planetary surface. Which set is correct?
ⓐ. \(V_s=-12b,\quad v_{e,s}=\sqrt{24b},\quad T/T_0=\sqrt3\)
ⓑ. \(V_s=-3b,\quad v_{e,s}=\sqrt{6b},\quad T/T_0=3\)
ⓒ. \(V_s=-6b,\quad v_{e,s}=\sqrt{6b},\quad T/T_0=9\)
ⓓ. \(V_s=-6b,\quad v_{e,s}=\sqrt{12b},\quad T/T_0=3\sqrt3\)
Correct Answer: \(V_s=-6b,\quad v_{e,s}=\sqrt{12b},\quad T/T_0=3\sqrt3\)
Explanation: \(\textbf{Translate altitude into orbital radius:}\)
The satellite is at altitude
\[
h=2R
\]
Therefore,
\[
r=R+h=3R
\]
\(\textbf{Use the given binding energy per unit mass:}\)
For a circular satellite,
\[
b=\frac{GM}{2r}
\]
At
\[
r=3R,
\]
\[
b=\frac{GM}{6R}
\]
Hence,
\[
\frac{GM}{R}=6b
\]
\(\textbf{Recover the surface gravitational potential:}\)
\[
V_s=-\frac{GM}{R}
\]
Therefore,
\[
V_s=-6b
\]
\(\textbf{Recover the surface escape speed:}\)
\[
v_{e,s}^2=\frac{2GM}{R}
\]
Thus,
\[
v_{e,s}^2=12b,
\]
so
\[
v_{e,s}=\sqrt{12b}
\]
\(\textbf{Compare orbital periods:}\)
For the same central mass,
\[
T\propto r^{3/2}
\]
Therefore,
\[
\frac{T}{T_0}
=
\left(\frac{3R}{R}\right)^{3/2}
\]
Hence,
\[
\frac{T}{T_0}=3\sqrt3
\]
\(\textbf{Final answer:}\)
\[
\boxed{V_s=-6b,\qquad
v_{e,s}=\sqrt{12b},\qquad
\frac{T}{T_0}=3\sqrt3}
\]
511. Two point masses A and B are fixed \(9a\) apart. Mass A is \(4m\), while mass B is \(\lambda m\). A test mass placed between them at a point \(6a\) from A experiences zero net gravitational force. The acceleration of A due to B is measured to be \(\alpha\). Which pair correctly gives \(\lambda\) and \(G\)?
ⓐ. \(\lambda=1,\quad G=\dfrac{81\alpha a^2}{m}\)
ⓑ. \(\lambda=4,\quad G=\dfrac{81\alpha a^2}{m}\)
ⓒ. \(\lambda=1,\quad G=\dfrac{9\alpha a^2}{m}\)
ⓓ. \(\lambda=\dfrac14,\quad G=\dfrac{81\alpha a^2}{m}\)
Correct Answer: \(\lambda=1,\quad G=\dfrac{81\alpha a^2}{m}\)
Explanation: \(\textbf{Use the neutral point to infer the unknown source mass:}\)
The test point is \(6a\) from A and therefore
\[
9a-6a=3a
\]
from B.
Zero resultant requires the two gravitational attractions to have equal magnitudes:
\[
G\frac{(4m)m_t}{(6a)^2}
=
G\frac{(\lambda m)m_t}{(3a)^2}
\]
Canceling common factors,
\[
\frac4{36}=\frac{\lambda}{9}
\]
Hence,
\[
\lambda=1
\]
\(\textbf{Use the mutual acceleration as an independent measurement:}\)
Mass B is therefore \(m\).
The acceleration of A due to B is
\[
\alpha=\frac{Gm}{(9a)^2}
\]
Thus,
\[
\alpha=\frac{Gm}{81a^2}
\]
Rearranging,
\[
G=\frac{81\alpha a^2}{m}
\]
The test mass used to locate the neutral point cancels completely and is not needed in the final result.
\(\textbf{Final answer:}\)
\[
\boxed{\lambda=1,\qquad G=\frac{81\alpha a^2}{m}}
\]
512. Two homogeneous spherical bodies A and B have radii \(R\) and \(2R\), respectively. The density of A is \(\rho\), while that of B is \(\rho_B\). Their surface gap \(s\) is varied, and the mutual gravitational force \(F\) is plotted against
\[
X=\frac{1}{(s+3R)^2}.
\]
The graph of \(F\) against \(X\) is a straight line through the origin with slope \(S\).
Which pair correctly gives \(\rho_B\) and the force \(F_c\) when the two bodies just touch?
ⓐ. \(\rho_B=\dfrac{9S}{128\pi^2G\rho R^6},\quad F_c=\dfrac{S}{4R^2}\)
ⓑ. \(\rho_B=\dfrac{9S}{16\pi^2G\rho R^6},\quad F_c=\dfrac{S}{3R^2}\)
ⓒ. \(\rho_B=\dfrac{128\pi^2G\rho R^6}{9S},\quad F_c=\dfrac{S}{9R^2}\)
ⓓ. \(\rho_B=\dfrac{9S}{128\pi^2G\rho R^6},\quad F_c=\dfrac{S}{9R^2}\)
Correct Answer: \(\rho_B=\dfrac{9S}{128\pi^2G\rho R^6},\quad F_c=\dfrac{S}{9R^2}\)
Explanation: \(\textbf{Interpret the graph variable geometrically:}\)
The centre-to-centre separation is
\[
d=s+R+2R=s+3R
\]
For external spherical bodies,
\[
F=G\frac{M_AM_B}{d^2}
\]
Since
\[
X=\frac1{d^2},
\]
the graph relation is
\[
F=(GM_AM_B)X
\]
Therefore,
\[
S=GM_AM_B
\]
\(\textbf{Express the two masses using density and radius:}\)
For A,
\[
M_A=\frac43\pi\rho R^3
\]
For B,
\[
M_B=\frac43\pi\rho_B(2R)^3
=\frac{32}{3}\pi\rho_BR^3
\]
Hence,
\[
S
=
G\left(\frac43\pi\rho R^3\right)
\left(\frac{32}{3}\pi\rho_BR^3\right)
\]
Therefore,
\[
S=\frac{128}{9}\pi^2G\rho\rho_BR^6
\]
Thus,
\[
\rho_B=\frac{9S}{128\pi^2G\rho R^6}
\]
\(\textbf{Evaluate the contact force:}\)
At contact,
\[
s=0,
\]
so
\[
d=3R
\]
Using the graph slope,
\[
F_c=\frac{S}{(3R)^2}
\]
Hence,
\[
F_c=\frac{S}{9R^2}
\]
\(\textbf{Final answer:}\)
\[
\boxed{\rho_B=\frac{9S}{128\pi^2G\rho R^6},\qquad
F_c=\frac{S}{9R^2}}
\]
513. Point masses \(M\) and \(4M\) are fixed at \(x=0\) and \(x=3a\), respectively. A negligible test body is placed at the point P between them where the net gravitational force is zero. Taking gravitational potential to be zero at infinity, what is the minimum speed that must be given to the body at P so that it can escape from the combined gravitational field of the two fixed masses?
ⓐ. \(\sqrt{\dfrac{2GM}{a}}\)
ⓑ. \(\sqrt{\dfrac{3GM}{a}}\)
ⓒ. \(\sqrt{\dfrac{6GM}{a}}\)
ⓓ. \(3\sqrt{\dfrac{GM}{a}}\)
Correct Answer: \(\sqrt{\dfrac{6GM}{a}}\)
Explanation: \(\textbf{Locate the zero-force point first:}\)
Let P be at distance \(x\) from mass \(M\).
Its distance from mass \(4M\) is
\[
3a-x
\]
At P,
\[
G\frac{M}{x^2}
=
G\frac{4M}{(3a-x)^2}
\]
Taking positive square roots,
\[
\frac1x=\frac2{3a-x}
\]
Hence,
\[
3a-x=2x,
\]
so
\[
x=a
\]
Thus the distances from P are \(a\) and \(2a\).
\(\textbf{Do not confuse zero field with zero potential:}\)
The potential at P is the scalar sum
\[
V_P=-\frac{GM}{a}-\frac{G(4M)}{2a}
\]
Therefore,
\[
V_P=-\frac{3GM}{a}
\]
For minimum escape from a fixed gravitational configuration,
\[
\frac12v_e^2+V_P=0
\]
Hence,
\[
\frac12v_e^2=\frac{3GM}{a}
\]
Therefore,
\[
v_e^2=\frac{6GM}{a}
\]
\(\textbf{Final answer:}\)
\[
v_e=\sqrt{\frac{6GM}{a}}
\]
The force at P is zero, but the body is still gravitationally bound because the potential there is negative.
514. At an external point P around an isolated spherical mass, the conventional gravitational potential with \(V_\infty=0\) is
\[
V_P=-40\,\text{MJ kg}^{-1}.
\]
A new reference is chosen by adding \(50\,\text{MJ kg}^{-1}\) to the potential everywhere. Thus
\[
V_P'=+10\,\text{MJ kg}^{-1},
\qquad
V_\infty'=+50\,\text{MJ kg}^{-1}.
\]
What is the escape speed from P?
ⓐ. \(10\,\text{km s}^{-1}\)
ⓑ. \(2\sqrt5\,\text{km s}^{-1}\)
ⓒ. \(4\sqrt5\,\text{km s}^{-1}\)
ⓓ. Escape is impossible because \(V_P'\gt 0\)
Correct Answer: \(4\sqrt5\,\text{km s}^{-1}\)
Explanation: Changing the zero of potential cannot change any physical speed.
The escape condition depends on the potential difference between the starting point and infinity.
With the shifted reference,
\[
\frac12v_e^2+V_P'=V_\infty'
\]
Therefore,
\[
\frac12v_e^2
=
V_\infty'-V_P'
\]
Substituting,
\[
\frac12v_e^2
=
50-10
=
40\,\text{MJ kg}^{-1}
\]
Hence,
\[
v_e^2=80\,\text{MJ kg}^{-1}
\]
Since
\[
1\,\text{MJ kg}^{-1}
=
1\,(\text{km s}^{-1})^2,
\]
\[
v_e=\sqrt{80}\,\text{km s}^{-1}
\]
Thus,
\[
v_e=4\sqrt5\,\text{km s}^{-1}
\]
Using \(v_e^2=-2V'\) directly would be invalid because the shifted potential no longer has zero value at infinity.
\(\textbf{Final answer:}\)
\[
\boxed{v_e=4\sqrt5\,\text{km s}^{-1}}
\]
515. A body of mass \(m\) moves around an isolated spherical mass \(M\) along the following closed path:
P \(\rightarrow\) Q: radially outward from \(R\) to \(2R\);
Q \(\rightarrow\) S: along a semicircular arc of radius \(2R\);
S \(\rightarrow\) T: radially inward from \(2R\) to \(R\);
T \(\rightarrow\) P: along an arc of radius \(R\).
Which set gives the work done by gravity on the four segments and on the complete closed path?
ⓐ. \(-\dfrac{GMm}{2R},\,0,\,+\dfrac{GMm}{2R},\,0;\quad W_{\text{closed}}=0\)
ⓑ. \(+\dfrac{GMm}{2R},\,0,\,-\dfrac{GMm}{2R},\,0;\quad W_{\text{closed}}=0\)
ⓒ. \(-\dfrac{GMm}{2R},\,-\dfrac{GMm}{2R},\,+\dfrac{GMm}{2R},\,+\dfrac{GMm}{2R};\quad W_{\text{closed}}=0\)
ⓓ. \(-\dfrac{GMm}{2R},\,0,\,+\dfrac{GMm}{2R},\,0;\quad W_{\text{closed}}=-\dfrac{GMm}{2R}\)
Correct Answer: \(-\dfrac{GMm}{2R},\,0,\,+\dfrac{GMm}{2R},\,0;\quad W_{\text{closed}}=0\)
Explanation: \(\textbf{Radial outward segment P }\rightarrow\textbf{ Q:}\)
The gravitational potential energy changes from
\[
-\frac{GMm}{R}
\]
to
\[
-\frac{GMm}{2R}
\]
Therefore,
\[
\Delta U_{PQ}
=
+\frac{GMm}{2R}
\]
Since
\[
W_g=-\Delta U,
\]
\[
W_{PQ}=-\frac{GMm}{2R}
\]
\(\textbf{Constant-radius arc Q }\rightarrow\textbf{ S:}\)
The gravitational force is radial while the infinitesimal displacement is tangential.
Hence,
\[
W_{QS}=0
\]
\(\textbf{Radial return S }\rightarrow\textbf{ T:}\)
The potential-energy change is the negative of the first radial change.
Therefore,
\[
W_{ST}=+\frac{GMm}{2R}
\]
\(\textbf{Final constant-radius arc T }\rightarrow\textbf{ P:}\)
Again,
\[
W_{TP}=0
\]
Thus,
\[
W_{\text{closed}}
=
-\frac{GMm}{2R}
+0
+\frac{GMm}{2R}
+0
=0
\]
The result is consistent with gravity being a conservative force.
\(\textbf{Final answer:}\)
\[
\boxed{W_{PQ}=-\frac{GMm}{2R},\quad
W_{QS}=0,\quad
W_{ST}=+\frac{GMm}{2R},\quad
W_{TP}=0,\quad
W_{\text{closed}}=0}
\]
516. At a point P, three fixed source masses produce the following gravitational-potential contributions, where \(\Phi_0\gt 0\) has unit \(\text{J kg}^{-1}\).
| Source | Potential contribution at P |
|---|
| A | \(-2\Phi_0\) |
| B | \(-3\Phi_0\) |
| C | \(-\Phi_0\) |
A small body is launched from P with speed satisfying
\[
v^2=10\Phi_0.
\]
Taking total potential to vanish at infinity, which statement is correct?
ⓐ. Unbound; \(v_\infty=\sqrt{2\Phi_0}\)
ⓑ. At the escape threshold; additional energy per unit mass \(=0\)
ⓒ. Bound; additional energy per unit mass \(=\Phi_0\)
ⓓ. Bound; additional energy per unit mass \(=6\Phi_0\)
Correct Answer: Bound; additional energy per unit mass \(=\Phi_0\)
Explanation: \(\textbf{Combine the source potentials as scalars:}\)
The total gravitational potential at P is
\[
V_P
=
-2\Phi_0-3\Phi_0-\Phi_0
\]
Hence,
\[
V_P=-6\Phi_0
\]
\(\textbf{Evaluate the launch kinetic energy per unit mass:}\)
\[
\frac Km=\frac12v^2
\]
Using
\[
v^2=10\Phi_0,
\]
\[
\frac Km=5\Phi_0
\]
Therefore the conserved total mechanical energy per unit mass is
\[
\varepsilon
=
\frac Km+V_P
\]
Thus,
\[
\varepsilon
=
5\Phi_0-6\Phi_0
=
-\Phi_0
\]
The negative result means the body is below the infinity-zero escape threshold.
To raise the conserved specific energy to zero requires an additional
\[
\Delta\varepsilon=\Phi_0
\]
The individual gravitational forces need not be calculated because escape classification here follows directly from scalar potential and total energy.
\(\textbf{Final answer:}\) The body is bound and requires an additional energy \(\Phi_0\) per unit mass.
517. A planet moves in an elliptical orbit around a star. Its perihelion and aphelion centre-distances are \(r\) and \(3r\), respectively, and its orbital period is \(T\). What are the gravitational potential due to the star and the local escape speed at the aphelion point?
ⓐ. \(V_a=-\dfrac{8\pi^2r^2}{3T^2},\quad v_{e,a}=\dfrac{4\pi r}{3T}\)
ⓑ. \(V_a=-\dfrac{16\pi^2r^2}{3T^2},\quad v_{e,a}=\dfrac{4\pi r}{\sqrt3\,T}\)
ⓒ. \(V_a=-\dfrac{32\pi^2r^2}{T^2},\quad v_{e,a}=\dfrac{8\pi r}{T}\)
ⓓ. \(V_a=-\dfrac{32\pi^2r^2}{3T^2},\quad v_{e,a}=\dfrac{8\pi r}{\sqrt3\,T}\)
Correct Answer: \(V_a=-\dfrac{32\pi^2r^2}{3T^2},\quad v_{e,a}=\dfrac{8\pi r}{\sqrt3\,T}\)
Explanation: \(\textbf{Extract the semi-major axis from the elliptical geometry:}\)
\[
a=\frac{r_p+r_a}{2}
\]
Therefore,
\[
a=\frac{r+3r}{2}=2r
\]
\(\textbf{Use the orbital period to infer the stellar gravitational parameter:}\)
Kepler's third-law form is
\[
T^2=\frac{4\pi^2a^3}{GM}
\]
Hence,
\[
GM=\frac{4\pi^2(2r)^3}{T^2}
\]
Therefore,
\[
GM=\frac{32\pi^2r^3}{T^2}
\]
\(\textbf{Evaluate the gravitational potential at aphelion:}\)
The aphelion distance is
\[
r_a=3r
\]
Thus,
\[
V_a=-\frac{GM}{3r}
\]
Substituting the inferred \(GM\),
\[
V_a
=
-\frac{32\pi^2r^2}{3T^2}
\]
\(\textbf{Use the local potential for escape speed:}\)
With zero potential at infinity,
\[
v_{e,a}^2=-2V_a
\]
Therefore,
\[
v_{e,a}^2
=
\frac{64\pi^2r^2}{3T^2}
\]
Hence,
\[
v_{e,a}
=
\frac{8\pi r}{\sqrt3\,T}
\]
\(\textbf{Final answer:}\)
\[
\boxed{V_a=-\frac{32\pi^2r^2}{3T^2},\qquad
v_{e,a}=\frac{8\pi r}{\sqrt3\,T}}
\]
518. In a gravitational-energy calculation, the zero of potential energy is deliberately shifted so that the potential energy of a particular test body at infinity is
\[
U_\infty=+12\,\text{J}.
\]
The body's conserved total mechanical energy is
\[
E=+5\,\text{J}.
\]
Which conclusion is correct?
ⓐ. The body is unbound because its total mechanical energy is positive
ⓑ. The body is exactly at the escape threshold because \(E\gt 0\)
ⓒ. The body cannot reach infinity without at least \(7\,\text{J}\) of additional energy
ⓓ. The body reaches infinity with kinetic energy \(17\,\text{J}\)
Correct Answer: The body cannot reach infinity without at least \(7\,\text{J}\) of additional energy
Explanation: The sign of a numerical total energy has physical meaning only after the reference level has been specified.
Here the chosen convention gives
\[
U_\infty=12\,\text{J},
\]
not zero.
If the body were to reach infinity, its kinetic energy there would have to satisfy
\[
E=K_\infty+U_\infty
\]
Therefore,
\[
K_\infty=E-U_\infty
\]
Substitution gives
\[
K_\infty=5-12=-7\,\text{J}
\]
A negative kinetic energy is impossible.
Thus the present energy is insufficient for the body to reach infinity.
The limiting escape state in this reference convention has
\[
E_{\text{threshold}}=U_\infty=12\,\text{J}
\]
Hence the minimum additional energy required is
\[
12-5=7\,\text{J}
\]
This is the same physical binding-energy requirement that would have been obtained using the conventional zero at infinity.
\(\textbf{Final answer:}\)
\[
\Delta E_{\min}=7\,\text{J}
\]
519. Two negligible bodies are launched from the same external point of an isolated spherical planet. Each is given a speed exactly equal to the local escape speed. Body P is projected radially outward, while body Q is projected tangentially. Atmosphere and all other interactions are neglected, and neither trajectory intersects the planet after launch. Which statement is correct?
ⓐ. Only P escapes because an escape trajectory must initially be radial
ⓑ. Only Q escapes because any tangential launch produces an unbound orbit
ⓒ. Both reach the zero-energy escape threshold; the launch direction need not be radial
ⓓ. Neither escapes because the launch speed must be greater than \(v_e\)
Correct Answer: Both reach the zero-energy escape threshold; the launch direction need not be radial
Explanation: At the common launch point, both bodies have the same gravitational potential energy.
They are also given the same speed magnitude.
Therefore they have the same kinetic energy and the same total mechanical energy.
For the local escape speed,
\[
\frac12mv_e^2-\frac{GMm}{r}=0.
\]
Thus each body has
\[
E=0.
\]
The radial launch has zero angular momentum, whereas the tangential launch has nonzero angular momentum, so their trajectories are different.
However, their escape classification is determined by the same zero-energy threshold.
The tangentially launched body is not in a circular orbit because
\[
v_e=\sqrt2\,v_o
\]
at the same radius.
Under the stated ideal conditions, both can reach arbitrarily large distance with limiting speed tending to zero.
Thus the customary term “escape velocity” refers here to the magnitude of the minimum required initial velocity, i.e. the escape speed.
520. Two equal fixed point masses \(M\) are at \((-a,0)\) and \((a,0)\). A negligible body is at
\[
P=(0,\sqrt3\,a).
\]
A student replaces the two source masses by a single mass \(2M\) at their centre of mass before calculating the escape speed from P. Which statement is correct?
ⓐ. The replacement is exact, and \(v_e=\sqrt{\dfrac{4GM}{\sqrt3\,a}}\)
ⓑ. The replacement is not exact; scalar superposition gives \(v_e=\sqrt{\dfrac{2GM}{a}}\)
ⓒ. The replacement is not exact; the net potential at P is zero by symmetry
ⓓ. The replacement is exact because the net horizontal gravitational-force components cancel
Correct Answer: The replacement is not exact; scalar superposition gives \(v_e=\sqrt{\dfrac{2GM}{a}}\)
Explanation: \(\textbf{Find the true distance from P to each source:}\)
For either mass,
\[
d=\sqrt{a^2+(\sqrt3a)^2}
\]
Hence,
\[
d=2a
\]
\(\textbf{Use scalar potential superposition:}\)
Each source contributes
\[
-\frac{GM}{2a}
\]
Therefore,
\[
V_P
=
-\frac{GM}{2a}
-\frac{GM}{2a}
\]
Thus,
\[
V_P=-\frac{GM}{a}
\]
The minimum escape condition is
\[
\frac12v_e^2+V_P=0
\]
Hence,
\[
v_e^2=\frac{2GM}{a}
\]
Therefore,
\[
v_e=\sqrt{\frac{2GM}{a}}
\]
The two discrete masses cannot generally be replaced exactly by a point mass \(2M\) at their centre of mass for a nearby field or potential calculation.
Cancellation of some force components by symmetry does not justify that replacement.
Only special source symmetries, such as an external point relative to a spherically symmetric body, permit the exact point-mass substitution used elsewhere in gravitation.
\(\textbf{Final answer:}\)
\[
v_e=\sqrt{\frac{2GM}{a}}
\]
