301. Four equal point masses \(m\) are placed at the vertices of a rectangle of sides \(3a\) and \(4a\). The entire rectangle is then enlarged so that every linear dimension doubles, while each mass is simultaneously changed from \(m\) to \(2m\). Let \(K=Gm^2/a\). What are the initial potential energy and the change in gravitational potential energy \(\Delta U=U_f-U_i\)?
ⓐ. \(U_i=-\dfrac{47K}{30},\quad \Delta U=+\dfrac{47K}{30}\)
ⓑ. \(U_i=-\dfrac{47K}{15},\quad \Delta U=-\dfrac{47K}{30}\)
ⓒ. \(U_i=-\dfrac{47K}{30},\quad \Delta U=-\dfrac{47K}{30}\)
ⓓ. \(U_i=-\dfrac{47K}{30},\quad \Delta U=-\dfrac{47K}{15}\)
Correct Answer: \(U_i=-\dfrac{47K}{30},\quad \Delta U=-\dfrac{47K}{30}\)
Explanation: \(\textbf{Energy of the initial rectangle:}\)
There are two pairs separated by \(3a\), two pairs separated by \(4a\), and two diagonal pairs separated by \(5a\).
Therefore,
\[
U_i
=
-Gm^2
\left(
\frac{2}{3a}
+\frac{2}{4a}
+\frac{2}{5a}
\right)
\]
Using \(K=Gm^2/a\),
\[
U_i
=
-K
\left(
\frac23+\frac12+\frac25
\right)
\]
With common denominator \(30\),
\[
\frac23+\frac12+\frac25
=
\frac{20+15+12}{30}
=
\frac{47}{30}
\]
Thus,
\[
U_i=-\frac{47K}{30}
\]
\(\textbf{Scale every pair interaction:}\)
Each mass doubles, so each mass product increases by factor
\[
2\times2=4
\]
Every separation doubles, giving an additional factor \(1/2\).
Hence every pair-energy term is multiplied by
\[
\frac42=2
\]
Therefore,
\[
U_f=2U_i=-\frac{47K}{15}
\]
\(\textbf{Calculate the energy change:}\)
\[
\Delta U=U_f-U_i
\]
Thus,
\[
\Delta U
=
-\frac{47K}{15}
+\frac{47K}{30}
=
-\frac{47K}{30}
\]
\(\textbf{Final answer:}\)
\[
\boxed{U_i=-\frac{47K}{30},\qquad
\Delta U=-\frac{47K}{30}}
\]
302. Point masses \(M,2M,4M\) are fixed at the vertices of an equilateral triangle of side \(a\). The mass \(2M\) is removed quasistatically to infinity while the other two remain fixed. What are the initial total gravitational potential energy and the external work required for removal?
ⓐ. \(U_i=-\dfrac{14GM^2}{a},\quad W_{\text{ext}}=\dfrac{10GM^2}{a}\)
ⓑ. \(U_i=-\dfrac{10GM^2}{a},\quad W_{\text{ext}}=\dfrac{14GM^2}{a}\)
ⓒ. \(U_i=-\dfrac{14GM^2}{a},\quad W_{\text{ext}}=\dfrac{4GM^2}{a}\)
ⓓ. \(U_i=-\dfrac{8GM^2}{a},\quad W_{\text{ext}}=\dfrac{10GM^2}{a}\)
Correct Answer: \(U_i=-\dfrac{14GM^2}{a},\quad W_{\text{ext}}=\dfrac{10GM^2}{a}\)
Explanation: \(\textbf{Sum the three initial pair interactions:}\)
For pair \(M,2M\),
\[
U_{12}=-\frac{2GM^2}{a}
\]
For pair \(M,4M\),
\[
U_{14}=-\frac{4GM^2}{a}
\]
For pair \(2M,4M\),
\[
U_{24}=-\frac{8GM^2}{a}
\]
Therefore,
\[
U_i
=
-\frac{(2+4+8)GM^2}{a}
\]
Hence,
\[
U_i=-\frac{14GM^2}{a}
\]
\(\textbf{Energy after removing }2M\textbf{:}\)
Only the \(M\)-\(4M\) pair remains:
\[
U_f=-\frac{4GM^2}{a}
\]
Thus,
\[
W_{\text{ext}}=U_f-U_i
\]
Therefore,
\[
W_{\text{ext}}
=
-\frac{4GM^2}{a}
+\frac{14GM^2}{a}
\]
Hence,
\[
W_{\text{ext}}=\frac{10GM^2}{a}
\]
The required work equals the magnitude of the interaction energy that the removed mass originally had with the other two masses.
\(\textbf{Final answer:}\)
\[
\boxed{U_i=-\frac{14GM^2}{a},\qquad
W_{\text{ext}}=\frac{10GM^2}{a}}
\]
303. A four-mass gravitational system has total potential energy \(-20\,\text{J}\). At the positions of three of the masses, the quantities \(m_iV_i\), where \(V_i\) is due only to the other masses, are
\[
-8\,\text{J},\quad -10\,\text{J},\quad -6\,\text{J}.
\]
What must \(m_4V_4\) be? If a fifth mass is then inserted quasistatically and its interaction energy with the original four-mass system is \(-5\,\text{J}\), what is the new total potential energy?
ⓐ. \(-8\,\text{J},\ -25\,\text{J}\)
ⓑ. \(-16\,\text{J},\ -25\,\text{J}\)
ⓒ. \(-16\,\text{J},\ -45\,\text{J}\)
ⓓ. \(-20\,\text{J},\ -30\,\text{J}\)
Correct Answer: \(-16\,\text{J},\ -25\,\text{J}\)
Explanation: \(\textbf{Use the double-counting relation:}\)
For the original four-mass system,
\[
U=\frac12\sum_i m_iV_i
\]
Given
\[
U=-20\,\text{J},
\]
we must have
\[
\sum_i m_iV_i=-40\,\text{J}
\]
The first three contributions sum to
\[
-8-10-6=-24\,\text{J}
\]
Therefore,
\[
m_4V_4=-40-(-24)
\]
Hence,
\[
m_4V_4=-16\,\text{J}
\]
\(\textbf{Insert the fifth mass:}\)
The stated interaction energy between the new mass and the existing four-mass system is
\[
\Delta U=-5\,\text{J}
\]
This is added once to the original system energy:
\[
U_{\text{new}}
=
-20-5
\]
Therefore,
\[
U_{\text{new}}=-25\,\text{J}
\]
No extra factor \(1/2\) is introduced when a new mass's total interaction with the pre-existing system has already been specified.
\(\textbf{Final answer:}\)
\[
\boxed{m_4V_4=-16\,\text{J},\qquad
U_{\text{new}}=-25\,\text{J}}
\]
304. A body is raised quasistatically above a spherical planet in two successive equal-height stages: from centre-distance \(R\) to \(R+h\), and then from \(R+h\) to \(R+2h\). The external work in the first stage is twice that in the second stage. What are \(h/R\) and the total exact increase in potential energy after both stages?
ⓐ. \(\dfrac13,\ \dfrac25mgR\)
ⓑ. \(1,\ \dfrac23mgR\)
ⓒ. \(\dfrac14,\ \dfrac13mgR\)
ⓓ. \(\dfrac12,\ \dfrac12mgR\)
Correct Answer: \(\dfrac12,\ \dfrac12mgR\)
Explanation: \(\textbf{Write the two exact energy increments:}\)
For the first stage,
\[
W_1
=
GMm\left(\frac1R-\frac1{R+h}\right)
\]
Therefore,
\[
W_1
=
\frac{GMm\,h}{R(R+h)}
\]
For the second stage,
\[
W_2
=
GMm\left(\frac1{R+h}-\frac1{R+2h}\right)
\]
Hence,
\[
W_2
=
\frac{GMm\,h}{(R+h)(R+2h)}
\]
Thus,
\[
\frac{W_1}{W_2}
=
\frac{R+2h}{R}
\]
The condition \(W_1=2W_2\) gives
\[
\frac{R+2h}{R}=2
\]
Therefore,
\[
2h=R,
\]
so
\[
h=\frac R2
\]
After both stages the body is at
\[
R+2h=2R
\]
Hence the total increase is
\[
\Delta U
=
GMm\left(\frac1R-\frac1{2R}\right)
=
\frac{GMm}{2R}
\]
Using \(GM=gR^2\),
\[
\Delta U=\frac12mgR
\]
\(\textbf{Final answer:}\)
\[
\boxed{\frac hR=\frac12,\qquad
\Delta U=\frac12mgR}
\]
305. For a body raised from the surface of a spherical planet, define
\[
x=\frac hR,\qquad
y=\frac{\Delta U_{\text{exact}}}{mgR}.
\]
Which description of the \(y\)-versus-\(x\) graph is correct?
ⓐ. \(y=x\); a straight line increasing without bound
ⓑ. \(y=\dfrac{x}{1+x}\); an increasing curve approaching \(1\) asymptotically
ⓒ. \(y=\dfrac{1}{(1+x)^2}\); a decreasing curve approaching zero
ⓓ. \(y=\dfrac{x^2}{1+x}\); an increasing curve without an upper bound
Correct Answer: \(y=\dfrac{x}{1+x}\); an increasing curve approaching \(1\) asymptotically
Explanation: The exact energy increase from the surface to altitude \(h\) is
\[
\Delta U
=
GMm\left(\frac1R-\frac1{R+h}\right)
\]
This simplifies to
\[
\Delta U
=
\frac{GMm\,h}{R(R+h)}
\]
Using
\[
GM=gR^2,
\]
we obtain
\[
\Delta U
=
mgR\frac{h}{R+h}
\]
Divide by \(mgR\):
\[
y=\frac{h}{R+h}
\]
With
\[
x=\frac hR,
\]
this becomes
\[
y=\frac{x}{1+x}
\]
At \(x=0\),
\[
y=0
\]
As
\[
x\rightarrow\infty,
\]
\[
y\rightarrow1
\]
Thus the exact energy increase saturates at \(mgR\), unlike the straight-line \(mgh\) approximation.
\(\textbf{Final answer:}\)
\[
y=\frac{x}{1+x}
\]
306. For a body raised from the surface of a spherical planet, the amount by which the constant-\(g\) estimate \(mgh\) exceeds the exact potential-energy increase is found to be
\[
\frac16mgR.
\]
What are \(h/R\) and the exact increase in potential energy?
ⓐ. \(\dfrac13,\ \dfrac14mgR\)
ⓑ. \(1,\ \dfrac12mgR\)
ⓒ. \(\dfrac12,\ \dfrac13mgR\)
ⓓ. \(\dfrac23,\ \dfrac25mgR\)
Correct Answer: \(\dfrac12,\ \dfrac13mgR\)
Explanation: \(\textbf{Write the exact and approximate expressions:}\)
Let
\[
x=\frac hR
\]
The constant-\(g\) estimate is
\[
mgh=mgR\,x
\]
The exact increase is
\[
\Delta U_{\text{exact}}
=
mgR\frac{x}{1+x}
\]
Therefore the excess of the approximation is
\[
mgh-\Delta U_{\text{exact}}
=
mgR\left[
x-\frac{x}{1+x}
\right]
\]
Simplifying,
\[
mgh-\Delta U_{\text{exact}}
=
mgR\frac{x^2}{1+x}
\]
The given condition gives
\[
\frac{x^2}{1+x}=\frac16
\]
Hence,
\[
6x^2-x-1=0
\]
Factoring,
\[
(3x+1)(2x-1)=0
\]
The physically acceptable positive solution is
\[
x=\frac12
\]
Thus,
\[
\frac hR=\frac12
\]
The exact energy increase is
\[
\Delta U_{\text{exact}}
=
mgR\frac{1/2}{3/2}
\]
Therefore,
\[
\Delta U_{\text{exact}}=\frac13mgR
\]
\(\textbf{Final answer:}\)
\[
\boxed{\frac hR=\frac12,\qquad
\Delta U_{\text{exact}}=\frac13mgR}
\]
307. A body is raised through a known height \(H\) above the surface of a spherical planet. The exact increase in gravitational potential energy is found to be exactly \(\dfrac23mgH\), where \(g\) is the surface gravitational acceleration. What are the planetary radius and the gravitational acceleration at the final point?
ⓐ. \(R=2H,\quad g_H=\dfrac49g\)
ⓑ. \(R=3H,\quad g_H=\dfrac{9}{16}g\)
ⓒ. \(R=H,\quad g_H=\dfrac14g\)
ⓓ. \(R=2H,\quad g_H=\dfrac23g\)
Correct Answer: \(R=2H,\quad g_H=\dfrac49g\)
Explanation: \(\textbf{Use the exact-to-}mgh\textbf{ ratio:}\)
The exact energy increase is
\[
\Delta U_{\text{exact}}
=
mgH\frac{R}{R+H}
\]
The measured condition is
\[
\Delta U_{\text{exact}}
=
\frac23mgH
\]
Therefore,
\[
\frac{R}{R+H}=\frac23
\]
Cross-multiplying,
\[
3R=2R+2H
\]
Hence,
\[
R=2H
\]
\(\textbf{Find the local gravitational acceleration:}\)
At the final altitude,
\[
g_H
=
g\left(\frac{R}{R+H}\right)^2
\]
The ratio already found is
\[
\frac{R}{R+H}=\frac23
\]
Thus,
\[
g_H=g\left(\frac23\right)^2
\]
Therefore,
\[
g_H=\frac49g
\]
The same geometric ratio controls the energy correction linearly and the local gravitational acceleration quadratically.
\(\textbf{Final answer:}\)
\[
\boxed{R=2H,\qquad g_H=\frac49g}
\]
308. Uniform spherical planets A and B have the same mean density. Their radii are \(R\) and \(2R\), respectively. The same body is raised through the same physical height \(R\) above each surface. If the exact energy increases are \(\Delta U_A,\Delta U_B\), which statement is correct?
ⓐ. \(\Delta U_B/\Delta U_A=2\); \(mgh\) is twice the exact value for both planets
ⓑ. \(\Delta U_B/\Delta U_A=4\); \(mgh\) is \(3/2\) times exact for A and twice exact for B
ⓒ. \(\Delta U_B/\Delta U_A=8/3\); \(mgh\) is twice exact for A and \(3/2\) times exact for B
ⓓ. \(\Delta U_B/\Delta U_A=3/2\); \(mgh\) is exact for both planets
Correct Answer: \(\Delta U_B/\Delta U_A=8/3\); \(mgh\) is twice exact for A and \(3/2\) times exact for B
Explanation: \(\textbf{Compare the surface gravities:}\)
For uniform planets of equal density,
\[
g_s\propto R
\]
Therefore,
\[
g_B=2g_A
\]
\(\textbf{Exact energy increase for planet A:}\)
Here the height is
\[
h=R=R_A
\]
Thus,
\[
\Delta U_A
=
m g_A R\frac{R}{R+R}
\]
Hence,
\[
\Delta U_A=\frac12m g_A R
\]
\(\textbf{Exact energy increase for planet B:}\)
Planet B has radius \(2R\), but the same physical height is only \(R\).
Therefore,
\[
\Delta U_B
=
m(2g_A)R
\frac{2R}{3R}
\]
Hence,
\[
\Delta U_B=\frac43m g_A R
\]
Thus,
\[
\frac{\Delta U_B}{\Delta U_A}
=
\frac{4/3}{1/2}
=
\frac83
\]
\(\textbf{Compare the constant-}g\textbf{ estimates:}\)
For A,
\[
m g_Ah=m g_AR=2\Delta U_A
\]
For B,
\[
m g_Bh=2m g_AR=\frac32\Delta U_B
\]
The approximation is therefore poorer for the smaller planet because the same physical height is a larger fraction of its radius.
\(\textbf{Final answer:}\)
\[
\frac{\Delta U_B}{\Delta U_A}=\frac83
\]
309. Consider the exact increase in gravitational potential energy when a body is raised from the surface of a spherical planet through a positive height \(h\).
Statement I: \(\Delta U_{\text{exact}}\lt mgh\) for every \(h\gt 0\).
Statement II: The fractional overestimate of \(mgh\), measured relative to the exact value, is exactly \(h/R\).
Statement III: As \(h\rightarrow\infty\), \(\Delta U_{\text{exact}}\rightarrow mgR\).
Statement IV: As \(h\rightarrow\infty\), \(mgh\) also approaches the finite value \(mgR\).
Which statements are correct?
ⓐ. I and III only
ⓑ. I, II and III only
ⓒ. II and IV only
ⓓ. I, II, III and IV
Correct Answer: I, II and III only
Explanation: The exact energy increase is
\[
\Delta U_{\text{exact}}
=
mgh\frac{R}{R+h}.
\]
For every positive \(h\),
\[
\frac{R}{R+h}\lt 1.
\]
Therefore,
\[
\Delta U_{\text{exact}}\lt mgh,
\]
so Statement I is correct.
The ratio of the approximate to the exact value is
\[
\frac{mgh}{\Delta U_{\text{exact}}}
=
\frac{R+h}{R}
=
1+\frac hR.
\]
Hence the fractional overestimate relative to the exact value is
\[
\frac{mgh-\Delta U_{\text{exact}}}{\Delta U_{\text{exact}}}
=
\frac hR,
\]
so Statement II is correct.
The exact expression may also be written as
\[
\Delta U_{\text{exact}}
=
mgR\frac{h}{R+h}.
\]
As
\[
h\rightarrow\infty,
\]
the fraction tends to \(1\), giving
\[
\Delta U_{\text{exact}}\rightarrow mgR.
\]
Thus Statement III is correct.
By contrast,
\[
mgh\rightarrow\infty
\]
as \(h\rightarrow\infty\), so Statement IV is incorrect.
Therefore Statements I, II and III only are correct.
310. In the ideal gravitational model, the escape speed from a given point is
ⓐ. The speed for which gravitational force becomes zero at a finite distance
ⓑ. The minimum launch speed that reaches infinity with limiting speed zero
ⓒ. The speed required to maintain a circular orbit through the same point
ⓓ. The minimum launch speed, proportional to the mass of the escaping body
Correct Answer: The minimum launch speed that reaches infinity with limiting speed zero
Explanation: Escape speed is defined through the limiting condition for an unpowered body to reach infinity.
The body is given an initial speed and thereafter moves under gravity alone in the ideal model.
For the minimum possible escape speed, it reaches infinity with
\[
v_\infty=0.
\]
With gravitational potential energy chosen to be zero at infinity, the final mechanical energy in this limiting case is also zero.
Gravity does not suddenly disappear at some finite distance.
Instead, its strength decreases continuously with distance and tends toward zero only as the distance tends to infinity.
Escape speed is therefore an energy-threshold speed, not the distance at which gravity stops acting.
It must also not be confused with the circular orbital speed at the same radius.
311. Consider the following statements about escape from a spherical planet.
Statement I: For minimum escape, the body's speed tends to zero as its distance tends to infinity.
Statement II: Minimum escape occurs because gravitational acceleration becomes exactly zero at a finite distance from the planet.
Statement III: The standard escape-speed derivation neglects atmospheric resistance and rotational complications unless they are explicitly included.
Statement IV: A more massive projectile requires a larger ideal escape speed from the same point.
Which statements are correct?
ⓐ. I and III only
ⓑ. I, II and III only
ⓒ. II and IV only
ⓓ. I, III and IV only
Correct Answer: I and III only
Explanation: Statement I is correct because the minimum-escape case is the limiting trajectory for which
\[
v_\infty=0.
\]
Statement II is incorrect.
For an isolated spherical source,
\[
g(r)=\frac{GM}{r^2},
\]
which remains non-zero at every finite \(r\) and tends to zero only as \(r\to\infty\).
Statement III is correct because the standard formula is derived for an ideal gravitational model without atmospheric drag or rotational effects.
Statement IV is incorrect because the projectile mass cancels from the energy equation.
Hence different test masses launched from the same point have the same ideal escape speed.
Therefore only Statements I and III are correct.
312. A body of mass \(m\) is projected from centre-distance \(r\) from an isolated spherical mass \(M\). Which equation correctly represents the minimum-escape condition and leads to the correct escape speed?
ⓐ. \(\dfrac12mv_e^2+\dfrac{GMm}{r}=0,\quad v_e=\sqrt{\dfrac{2GM}{r}}\)
ⓑ. \(\dfrac12mv_e^2-\dfrac{GMm}{r}=0,\quad v_e=\sqrt{\dfrac{GM}{r}}\)
ⓒ. \(\dfrac12mv_e^2-\dfrac{GMm}{r}=0,\quad v_e=\sqrt{\dfrac{2GM}{r}}\)
ⓓ. \(\dfrac12mv_e^2-\dfrac{GMm}{r}=0,\quad v_e=\sqrt{\dfrac{2GMm}{r}}\)
Correct Answer: \(\dfrac12mv_e^2-\dfrac{GMm}{r}=0,\quad v_e=\sqrt{\dfrac{2GM}{r}}\)
Explanation: \(\textbf{Set the limiting total energy:}\)
At the launch point,
\[
K_i=\frac12mv_e^2,
\]
and
\[
U_i=-\frac{GMm}{r}
\]
For minimum escape, the body reaches infinity with zero final speed.
Since
\[
U(\infty)=0,
\]
the limiting final total energy is
\[
E_f=0
\]
Energy conservation therefore requires
\[
\frac12mv_e^2-\frac{GMm}{r}=0
\]
Canceling \(m\),
\[
\frac12v_e^2=\frac{GM}{r}
\]
Thus,
\[
v_e=\sqrt{\frac{2GM}{r}}
\]
The cancellation of \(m\) explains why ideal escape speed is independent of the escaping body's mass.
\(\textbf{Final answer:}\)
\[
v_e=\sqrt{\frac{2GM}{r}}
\]
313. Take Earth's radius as \(6.4\times10^6\,\text{m}\) and surface acceleration due to gravity as \(9.8\,\text{m s}^{-2}\). The ideal escape speed from Earth's surface is approximately
ⓐ. \(7.9\,\text{km s}^{-1}\)
ⓑ. \(9.8\,\text{km s}^{-1}\)
ⓒ. \(11.2\,\text{km s}^{-1}\)
ⓓ. \(15.8\,\text{km s}^{-1}\)
Correct Answer: \(11.2\,\text{km s}^{-1}\)
Explanation: \(\textbf{Use the surface escape-speed relation:}\)
\[
v_e=\sqrt{2gR}
\]
Substituting,
\[
v_e
=
\sqrt{2(9.8)(6.4\times10^6)}
\]
Therefore,
\[
v_e
=
\sqrt{1.2544\times10^8}
\]
Hence,
\[
v_e=1.12\times10^4\,\text{m s}^{-1}
\]
Converting to kilometres per second,
\[
v_e=11.2\,\text{km s}^{-1}
\]
This is the standard ideal surface escape speed obtained without including atmospheric drag or Earth's rotation.
\(\textbf{Final answer:}\)
\[
\boxed{v_e\approx11.2\,\text{km s}^{-1}}
\]
314. A spherical planet has surface acceleration due to gravity \(4.0\,\text{m s}^{-2}\) and surface escape speed \(8.0\,\text{km s}^{-1}\). Take \(G=6.67\times10^{-11}\,\text{N m}^2\text{kg}^{-2}\). What are its radius and mass?
ⓐ. \(R=4.0\times10^6\,\text{m},\quad M\approx9.6\times10^{23}\,\text{kg}\)
ⓑ. \(R=8.0\times10^6\,\text{m},\quad M\approx3.84\times10^{24}\,\text{kg}\)
ⓒ. \(R=8.0\times10^6\,\text{m},\quad M\approx7.68\times10^{24}\,\text{kg}\)
ⓓ. \(R=1.6\times10^7\,\text{m},\quad M\approx3.84\times10^{24}\,\text{kg}\)
Correct Answer: \(R=8.0\times10^6\,\text{m},\quad M\approx3.84\times10^{24}\,\text{kg}\)
Explanation: \(\textbf{Infer the planetary radius from }v_e\textbf{ and }g\textbf{:}\)
At the surface,
\[
v_e^2=2gR
\]
Therefore,
\[
R=\frac{v_e^2}{2g}
\]
Using
\[
v_e=8.0\times10^3\,\text{m s}^{-1},
\]
\[
R
=
\frac{(8.0\times10^3)^2}{2(4.0)}
\]
Thus,
\[
R
=
\frac{64\times10^6}{8}
=8.0\times10^6\,\text{m}
\]
\(\textbf{Recover the mass from the surface gravity:}\)
\[
g=\frac{GM}{R^2}
\]
Hence,
\[
M=\frac{gR^2}{G}
\]
Substitution gives
\[
M
=
\frac{4.0(8.0\times10^6)^2}
{6.67\times10^{-11}}
\]
Therefore,
\[
M\approx3.84\times10^{24}\,\text{kg}
\]
\(\textbf{Final answer:}\)
\[
\boxed{R=8.0\times10^6\,\text{m},\qquad
M\approx3.84\times10^{24}\,\text{kg}}
\]
315. A spherical planet of fixed mass \(M\) is compressed without loss of mass from radius \(R\) to \(R/4\). If its original surface escape speed, surface gravity and mean density are \(v_e,g,\rho\), respectively, what are their new values?
ⓐ. \(2v_e,\ 16g,\ 64\rho\)
ⓑ. \(4v_e,\ 16g,\ 16\rho\)
ⓒ. \(2v_e,\ 4g,\ 64\rho\)
ⓓ. \(4v_e,\ 4g,\ 16\rho\)
Correct Answer: \(2v_e,\ 16g,\ 64\rho\)
Explanation: \(\textbf{Escape-speed change:}\)
At fixed mass,
\[
v_e=\sqrt{\frac{2GM}{R}}
\]
Hence,
\[
v_e\propto R^{-1/2}
\]
Replacing \(R\) by \(R/4\),
\[
v_e'=v_e\sqrt4=2v_e
\]
\(\textbf{Surface-gravity change:}\)
\[
g=\frac{GM}{R^2}
\]
Therefore,
\[
g'=g\left(\frac{R}{R/4}\right)^2
\]
Thus,
\[
g'=16g
\]
\(\textbf{Density change:}\)
At fixed mass,
\[
\rho\propto\frac1{R^3}
\]
Hence,
\[
\rho'=\rho(4)^3=64\rho
\]
The three quantities therefore scale differently under the same compression.
\(\textbf{Final answer:}\)
\[
\boxed{v_e'=2v_e,\qquad g'=16g,\qquad \rho'=64\rho}
\]
316. A spherical planet has radius \(R\), surface gravity \(g\), and surface escape speed \(v_e\). What are the escape speed and local gravitational acceleration at altitude \(3R\) above its surface?
ⓐ. \(v_e/\sqrt2,\ g/4\)
ⓑ. \(v_e/4,\ g/16\)
ⓒ. \(v_e/2,\ g/4\)
ⓓ. \(v_e/2,\ g/16\)
Correct Answer: \(v_e/2,\ g/16\)
Explanation: \(\textbf{Convert altitude into centre-distance:}\)
At altitude
\[
h=3R,
\]
the distance from the planet's centre is
\[
r=R+h=4R
\]
\(\textbf{Escape speed at the new radius:}\)
For the same source mass,
\[
v_e(r)=\sqrt{\frac{2GM}{r}}
\]
Therefore,
\[
\frac{v_e(r)}{v_e(R)}
=
\sqrt{\frac{R}{4R}}
=\frac12
\]
Hence,
\[
v_e(r)=\frac{v_e}{2}
\]
\(\textbf{Local gravitational acceleration:}\)
\[
g(r)=\frac{GM}{r^2}
\]
Thus,
\[
\frac{g(r)}{g}
=
\left(\frac{R}{4R}\right)^2
=\frac1{16}
\]
Therefore,
\[
g(r)=\frac{g}{16}
\]
Escape speed falls as \(1/\sqrt r\), whereas local gravitational acceleration falls as \(1/r^2\).
\(\textbf{Final answer:}\)
\[
\boxed{v_e(r)=\frac{v_e}{2},\qquad g(r)=\frac{g}{16}}
\]
317. Uniform spherical planets A and B have the same mean density, and \(R_B=3R_A\). What are the ratios of their surface escape speeds, surface gravitational accelerations and masses?
ⓐ. \(3,\ 3,\ 27\)
ⓑ. \(\sqrt3,\ 3,\ 9\)
ⓒ. \(3,\ 9,\ 27\)
ⓓ. \(\sqrt3,\ \sqrt3,\ 27\)
Correct Answer: \(3,\ 3,\ 27\)
Explanation: \(\textbf{Mass scaling at equal density:}\)
For a uniform sphere,
\[
M\propto \rho R^3
\]
At equal density,
\[
\frac{M_B}{M_A}
=
\left(\frac{R_B}{R_A}\right)^3
=3^3=27
\]
\(\textbf{Surface-gravity scaling:}\)
For a uniform sphere,
\[
g=\frac43\pi G\rho R
\]
At equal density,
\[
g\propto R
\]
Therefore,
\[
\frac{g_B}{g_A}=3
\]
\(\textbf{Escape-speed scaling:}\)
Using
\[
v_e=\sqrt{2gR},
\]
\[
v_e\propto\sqrt{R\cdot R}=R
\]
when density is fixed.
Thus,
\[
\frac{v_{eB}}{v_{eA}}=3
\]
Equivalently, \(v_e\propto R\sqrt{\rho}\).
\(\textbf{Final answer:}\)
\[
\boxed{\frac{v_{eB}}{v_{eA}}=3,\qquad
\frac{g_B}{g_A}=3,\qquad
\frac{M_B}{M_A}=27}
\]
318. Planet B has surface gravitational acceleration one-fourth that of planet A, but its radius is four times that of A. If the planets are treated as uniform spheres, which pair is correct?
ⓐ. \(v_{eB}=v_{eA},\quad \rho_B/\rho_A=1/16\)
ⓑ. \(v_{eB}=2v_{eA},\quad \rho_B/\rho_A=1/4\)
ⓒ. \(v_{eB}=v_{eA}/2,\quad \rho_B/\rho_A=1/16\)
ⓓ. \(v_{eB}=2v_{eA},\quad \rho_B/\rho_A=1/16\)
Correct Answer: \(v_{eB}=v_{eA},\quad \rho_B/\rho_A=1/16\)
Explanation: \(\textbf{Compare the escape speeds using surface data:}\)
At the surface,
\[
v_e=\sqrt{2gR}
\]
Therefore,
\[
\frac{v_{eB}^2}{v_{eA}^2}
=
\frac{g_BR_B}{g_AR_A}
\]
Using
\[
g_B=\frac{g_A}{4},
\qquad
R_B=4R_A,
\]
we obtain
\[
\frac{v_{eB}^2}{v_{eA}^2}
=
\frac14(4)=1
\]
Hence,
\[
v_{eB}=v_{eA}
\]
\(\textbf{Infer the density ratio:}\)
For a uniform sphere,
\[
g=\frac43\pi G\rho R
\]
Thus,
\[
\rho\propto\frac gR
\]
Therefore,
\[
\frac{\rho_B}{\rho_A}
=
\frac{g_B}{g_A}
\frac{R_A}{R_B}
\]
Hence,
\[
\frac{\rho_B}{\rho_A}
=
\frac14\cdot\frac14
=\frac1{16}
\]
\(\textbf{Final answer:}\)
\[
\boxed{v_{eB}=v_{eA},\qquad
\frac{\rho_B}{\rho_A}=\frac1{16}}
\]
319. Two spherical planets A and B have equal surface escape speeds, while \(R_B=4R_A\). What are \(g_B/g_A\) and \(M_B/M_A\)?
ⓐ. \(1,\ 4\)
ⓑ. \(1/2,\ 8\)
ⓒ. \(1/4,\ 16\)
ⓓ. \(1/4,\ 4\)
Correct Answer: \(1/4,\ 4\)
Explanation: \(\textbf{Use equal escape speeds:}\)
At a planetary surface,
\[
v_e^2=2gR
\]
Equal escape speeds therefore imply
\[
g_AR_A=g_BR_B
\]
Given
\[
R_B=4R_A,
\]
we have
\[
g_A R_A
=
g_B(4R_A)
\]
Hence,
\[
\frac{g_B}{g_A}=\frac14
\]
\(\textbf{Infer the mass ratio:}\)
Surface gravity satisfies
\[
M=\frac{gR^2}{G}
\]
Therefore,
\[
\frac{M_B}{M_A}
=
\frac{g_B}{g_A}
\left(\frac{R_B}{R_A}\right)^2
\]
Substituting,
\[
\frac{M_B}{M_A}
=
\frac14(4)^2
\]
Thus,
\[
\frac{M_B}{M_A}=4
\]
Equal escape speed fixes the ratio \(M/R\), not the mass or radius separately.
\(\textbf{Final answer:}\)
\[
\boxed{\frac{g_B}{g_A}=\frac14,\qquad
\frac{M_B}{M_A}=4}
\]
320. Planet A has surface gravity \(g\), radius \(R\), and escape speed \(v\). The following data are reported for other spherical planets:
| Planet | Surface gravity | Radius | Reported escape speed |
|---|
| A | \(g\) | \(R\) | \(v\) |
| B | \(4g\) | \(R\) | \(2v\) |
| C | \(g\) | \(9R\) | \(2v\) |
| D | \(g/4\) | \(4R\) | \(v\) |
Which reported row is inconsistent?
ⓐ. A
ⓑ. B
ⓒ. C
ⓓ. D
Correct Answer: C
Explanation: For any spherical planet,
\[
v_e=\sqrt{2gR}.
\]
Therefore relative to planet A,
\[
\frac{v_e'}{v}
=
\sqrt{
\frac{g'R'}{gR}
}.
\]
For B,
\[
\frac{v_B}{v}
=
\sqrt{4}=2,
\]
so B is consistent.
For C,
\[
\frac{v_C}{v}
=
\sqrt{9}=3.
\]
The table reports \(2v\), so C is inconsistent.
For D,
\[
\frac{v_D}{v}
=
\sqrt{\frac14\cdot4}=1,
\]
so D is consistent.
Thus the correct escape speed for C should be
\[
3v.
\]
\(\textbf{Final answer:}\) Row C is inconsistent.
