101. At an unknown altitude \(h\) above a spherical planet of radius \(R\), the local acceleration due to gravity is \(4g/9\), where \(g\) is the surface value. A body at that position is then raised through an additional altitude \(R/2\). What is the new acceleration due to gravity?
ⓐ. \(\dfrac{4g}{25}\)
ⓑ. \(\dfrac{g}{9}\)
ⓒ. \(\dfrac{g}{4}\)
ⓓ. \(\dfrac{9g}{16}\)
Correct Answer: \(\dfrac{g}{4}\)
Explanation: \(\textbf{First infer the original altitude:}\)
The exact altitude relation is
\[
\frac{g_h}{g}
=
\left(\frac{R}{R+h}\right)^2
\]
Given
\[
\frac{g_h}{g}=\frac49,
\]
we have
\[
\frac{R}{R+h}=\frac23
\]
Therefore,
\[
h=\frac R2
\]
\(\textbf{Apply the second displacement:}\)
The body is raised by another \(R/2\), so its new altitude is
\[
h'=\frac R2+\frac R2=R
\]
Its new centre-distance is therefore
\[
R+h'=2R
\]
Hence,
\[
g'=\frac{GM}{(2R)^2}
\]
Using
\[
g=\frac{GM}{R^2},
\]
we obtain
\[
g'=\frac g4
\]
The first gravity measurement must be converted into an altitude before the second change can be applied.
\(\textbf{Final answer:}\)
\[
\boxed{g'=\frac g4}
\]
102. Planet A has mass \(M\) and radius \(R\). Planet B has mass \(8M\) and radius \(2R\). A body of mass \(3m\) is held at altitude \(R\) above A, while a body of mass \(m\) is held at altitude \(2R\) above B. If the local accelerations are \(g_A'\) and \(g_B'\), and the corresponding weights are \(W_A\) and \(W_B\), then
ⓐ. \(\dfrac{g_B'}{g_A'}=2,\quad\dfrac{W_B}{W_A}=2\)
ⓑ. \(\dfrac{g_B'}{g_A'}=4,\quad\dfrac{W_B}{W_A}=\dfrac43\)
ⓒ. \(\dfrac{g_B'}{g_A'}=\dfrac12,\quad\dfrac{W_B}{W_A}=\dfrac16\)
ⓓ. \(\dfrac{g_B'}{g_A'}=2,\quad\dfrac{W_B}{W_A}=\dfrac23\)
Correct Answer: \(\dfrac{g_B'}{g_A'}=2,\quad\dfrac{W_B}{W_A}=\dfrac23\)
Explanation: \(\textbf{Local gravity above planet A:}\)
The centre-distance is
\[
r_A=R+R=2R
\]
Therefore,
\[
g_A'=\frac{GM}{(2R)^2}
=\frac{GM}{4R^2}
\]
\(\textbf{Local gravity above planet B:}\)
Planet B has radius \(2R\), and the stated altitude is also \(2R\).
Thus,
\[
r_B=2R+2R=4R
\]
Hence,
\[
g_B'
=
\frac{G(8M)}{(4R)^2}
\]
So,
\[
g_B'=\frac{8GM}{16R^2}
=\frac{GM}{2R^2}
\]
Therefore,
\[
\frac{g_B'}{g_A'}
=
\frac{1/2}{1/4}=2
\]
\(\textbf{Compare the weights:}\)
\[
W_A=(3m)g_A',
\qquad
W_B=mg_B'
\]
Thus,
\[
\frac{W_B}{W_A}
=
\frac{m(2g_A')}{3mg_A'}
=\frac23
\]
\(\textbf{Final answer:}\)
\[
\boxed{\frac{g_B'}{g_A'}=2,\qquad
\frac{W_B}{W_A}=\frac23}
\]
103. For a spherical planet of known radius \(R\), measurements of gravitational acceleration are made at several altitudes \(h\).
The measured value \(g_h\) is plotted on the vertical axis against \(x=1/(R+h)^2\) on the horizontal axis. The graph is a straight line through the origin with slope \(S\).
Which relation correctly gives the planet's mass \(M\) and its surface gravity \(g\)?
ⓐ. \(M=\dfrac{G}{S},\quad g=SR^2\)
ⓑ. \(M=\dfrac{S}{G},\quad g=SR^2\)
ⓒ. \(M=\dfrac{S}{G},\quad g=\dfrac{S}{R^2}\)
ⓓ. \(M=\dfrac{S}{GR^2},\quad g=S\)
Correct Answer: \(M=\dfrac{S}{G},\quad g=\dfrac{S}{R^2}\)
Explanation: For an external point,
\[
g_h=\frac{GM}{(R+h)^2}
\]
With
\[
x=\frac{1}{(R+h)^2},
\]
the relation becomes
\[
g_h=(GM)x
\]
Thus the slope of the graph is
\[
S=GM
\]
Therefore the planetary mass is
\[
M=\frac{S}{G}
\]
At the surface,
\[
h=0,
\]
so the centre-distance is \(R\).
Hence,
\[
g=\frac{GM}{R^2}
\]
Since \(GM=S\),
\[
g=\frac{S}{R^2}
\]
The linear graph therefore gives the gravitational parameter \(GM\) directly through its slope.
\(\textbf{Final answer:}\)
\[
\boxed{M=\frac{S}{G},\qquad g=\frac{S}{R^2}}
\]
104. Consider the following statements about acceleration due to gravity above the surface of a spherical planet.
Statement I: For every positive altitude \(h\), \(g_h\lt g\), where \(g\) is the surface value.
Statement II: At a fixed altitude, a heavier test body experiences a larger value of \(g_h\).
Statement III: For a finite source mass, \(g_h\) approaches zero as \(h\to\infty\), but is not zero at any ordinary finite altitude.
Statement IV: Doubling the altitude always halves the local acceleration due to gravity.
Statement V: Outside the planet, gravitational acceleration follows an inverse-square dependence on distance from the planet's centre.
Which statements are correct?
ⓐ. I, II and V only
ⓑ. I, III and V only
ⓒ. II, III and IV only
ⓓ. I, III, IV and V only
Correct Answer: I, III and V only
Explanation: Statement I is correct because increasing altitude increases the centre-distance from \(R\) to \(R+h\), so the inverse-square gravitational acceleration decreases.
Statement II is incorrect because
\[
g_h=\frac{GM}{(R+h)^2}
\]
contains no test-body mass.
Statement III is correct because the denominator grows without bound as \(h\to\infty\), making \(g_h\to0\).
At any finite centre-distance from a non-zero spherical source mass, however, the expression remains non-zero.
Statement IV is incorrect because the dependence is not \(g_h\propto1/h\).
Altitude must first be added to the planetary radius, and the resulting centre-distance is then squared.
Statement V is correct because outside the spherical planet,
\[
g(r)=\frac{GM}{r^2}.
\]
Therefore Statements I, III and V are correct.
105. Assertion: The acceleration due to gravity decreases as a body is taken to increasing altitude above a spherical planet, although the universal gravitational constant \(G\) remains unchanged.
Reason: The planet's mass remains the same while the distance from its centre changes from \(R\) to \(R+h\), so the local acceleration is \(GM/(R+h)^2\).
ⓐ. 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: Both Assertion and Reason are correct, and Reason correctly explains Assertion
Explanation: The universal gravitational constant \(G\) is not a location-dependent quantity.
Taking a body upward does not alter the value of \(G\).
The source mass \(M\) of the planet is also unchanged in the stated situation.
What changes is the separation between the body and the planet's centre.
At the surface that separation is \(R\), whereas at altitude \(h\) it is \(R+h\).
Thus,
\[
g_h=\frac{GM}{(R+h)^2}.
\]
For every \(h\gt 0\),
\[
(R+h)^2\gt R^2,
\]
so \(g_h\lt g\).
The Reason therefore identifies the physical and mathematical cause of the decrease in gravitational acceleration.
Hence both Assertion and Reason are correct, and the Reason correctly explains the Assertion.
106. At a certain altitude above a spherical planet, the acceleration due to gravity is \(16\%\) of its surface value. What is the altitude in terms of the planetary radius \(R\)?
ⓐ. \(\dfrac{R}{2}\)
ⓑ. \(R\)
ⓒ. \(\dfrac{3R}{2}\)
ⓓ. \(2R\)
Correct Answer: \(\dfrac{3R}{2}\)
Explanation: \(\textbf{Translate the percentage into a ratio:}\)
\[
\frac{g_h}{g}=0.16=\frac{16}{100}=\frac4{25}
\]
The exact altitude relation is
\[
\frac{g_h}{g}
=
\left(\frac{R}{R+h}\right)^2
\]
Therefore,
\[
\left(\frac{R}{R+h}\right)^2=\frac4{25}
\]
Taking the positive square root,
\[
\frac{R}{R+h}=\frac25
\]
Hence,
\[
5R=2(R+h)
\]
Therefore,
\[
5R=2R+2h,
\]
so
\[
2h=3R
\]
Thus,
\[
h=\frac{3R}{2}
\]
The altitude is greater than the radius because reducing \(g\) to only \(16\%\) requires the centre-distance to increase to \(5R/2\).
\(\textbf{Final answer:}\)
\[
\boxed{h=\frac{3R}{2}}
\]
107. At an altitude of \(1000\,\text{km}\) above the surface of a spherical planet, the acceleration due to gravity is \(0.64\) times its surface value. What is the planet's radius?
ⓐ. \(1600\,\text{km}\)
ⓑ. \(2500\,\text{km}\)
ⓒ. \(3200\,\text{km}\)
ⓓ. \(4000\,\text{km}\)
Correct Answer: \(4000\,\text{km}\)
Explanation: \(\textbf{Use the exact gravity ratio:}\)
\[
\frac{g_h}{g}
=
\left(\frac{R}{R+h}\right)^2
\]
The given ratio is
\[
0.64=(0.8)^2
\]
Therefore,
\[
\frac{R}{R+h}=0.8
\]
With
\[
h=1000\,\text{km},
\]
we get
\[
\frac{R}{R+1000}=0.8
\]
Hence,
\[
R=0.8R+800
\]
Therefore,
\[
0.2R=800
\]
Thus,
\[
R=4000\,\text{km}
\]
A common error would be to treat \(1000\,\text{km}\) itself as the inverse-square distance instead of using \(R+h\).
\(\textbf{Final answer:}\)
\[
\boxed{R=4000\,\text{km}}
\]
108. Measurements of gravitational acceleration above a spherical planet give the following data.
| Altitude | Local acceleration due to gravity |
|---|
| \(1000\,\text{km}\) | \(4.00\,\text{m s}^{-2}\) |
| \(3000\,\text{km}\) | \(2.25\,\text{m s}^{-2}\) |
Assuming spherical symmetry, what are the planet's radius \(R\) and surface gravity \(g\)?
ⓐ. \(R=4000\,\text{km},\quad g=6.25\,\text{m s}^{-2}\)
ⓑ. \(R=5000\,\text{km},\quad g=5.76\,\text{m s}^{-2}\)
ⓒ. \(R=6000\,\text{km},\quad g=4.00\,\text{m s}^{-2}\)
ⓓ. \(R=5000\,\text{km},\quad g=4.80\,\text{m s}^{-2}\)
Correct Answer: \(R=5000\,\text{km},\quad g=5.76\,\text{m s}^{-2}\)
Explanation: \(\textbf{Use the two altitude measurements to infer the radius:}\)
For the same planet,
\[
g_h=\frac{GM}{(R+h)^2}
\]
Therefore,
\[
\frac{4.00}{2.25}
=
\left(\frac{R+3000}{R+1000}\right)^2
\]
Since
\[
\frac{4.00}{2.25}=\frac{16}{9},
\]
taking the positive square root gives
\[
\frac{R+3000}{R+1000}=\frac43
\]
Hence,
\[
3(R+3000)=4(R+1000)
\]
So,
\[
3R+9000=4R+4000
\]
Therefore,
\[
R=5000\,\text{km}
\]
\(\textbf{Recover the surface gravity:}\)
Using the first measurement,
\[
4.00
=
g\left(\frac{R}{R+1000}\right)^2
\]
With \(R=5000\,\text{km}\),
\[
4.00
=
g\left(\frac{5000}{6000}\right)^2
=
g\left(\frac56\right)^2
\]
Thus,
\[
4.00=g\frac{25}{36}
\]
Hence,
\[
g=4.00\frac{36}{25}=5.76\,\text{m s}^{-2}
\]
\(\textbf{Final answer:}\)
\[
\boxed{R=5000\,\text{km},\qquad g=5.76\,\text{m s}^{-2}}
\]
109. The same body has weight \(W_1\) at altitude \(h\) and weight \(W_2\) at altitude \(3h\) above a spherical planet of radius \(R\). It is found that
\[
W_1=4W_2.
\]
What are \(h\) and \(W_1/W_0\), where \(W_0\) is the body's surface weight?
ⓐ. \(h=R,\quad\dfrac{W_1}{W_0}=\dfrac14\)
ⓑ. \(h=\dfrac R2,\quad\dfrac{W_1}{W_0}=\dfrac49\)
ⓒ. \(h=R,\quad\dfrac{W_1}{W_0}=\dfrac19\)
ⓓ. \(h=2R,\quad\dfrac{W_1}{W_0}=\dfrac14\)
Correct Answer: \(h=R,\quad\dfrac{W_1}{W_0}=\dfrac14\)
Explanation: \(\textbf{Use the weight ratio as a gravity ratio:}\)
The same body is used, so
\[
\frac{W_1}{W_2}
=
\frac{g_h}{g_{3h}}
\]
Using the exact inverse-square relation,
\[
\frac{W_1}{W_2}
=
\left(\frac{R+3h}{R+h}\right)^2
\]
The measured ratio is
\[
4=
\left(\frac{R+3h}{R+h}\right)^2
\]
Taking the positive square root,
\[
2=\frac{R+3h}{R+h}
\]
Therefore,
\[
R+3h=2R+2h
\]
Hence,
\[
h=R
\]
\(\textbf{Compare the first position with the surface:}\)
At \(h=R\),
\[
\frac{W_1}{W_0}
=
\left(\frac{R}{R+R}\right)^2
\]
Thus,
\[
\frac{W_1}{W_0}=\frac14
\]
The first measurement is used to infer the altitude before its relation to surface weight can be found.
\(\textbf{Final answer:}\)
\[
\boxed{h=R,\qquad \frac{W_1}{W_0}=\frac14}
\]
110. A student calculates the acceleration due to gravity at altitude \(h=R/10\) in two ways. Method X uses the exact relation, while Method Y uses the small-altitude approximation. Which pair of results is correct?
ⓐ. Method X: \(0.80g\); Method Y: \(\dfrac{100}{121}g\)
ⓑ. Method X: \(\dfrac{10}{11}g\); Method Y: \(0.90g\)
ⓒ. Method X: \(\dfrac{100}{121}g\); Method Y: \(0.80g\)
ⓓ. Method X: \(\dfrac{121}{100}g\); Method Y: \(0.80g\)
Correct Answer: Method X: \(\dfrac{100}{121}g\); Method Y: \(0.80g\)
Explanation: \(\textbf{Exact calculation:}\)
The exact altitude dependence is
\[
g_h=g\left(\frac{R}{R+h}\right)^2
\]
For
\[
h=\frac R{10},
\]
\[
g_h
=
g\left(\frac{R}{R+R/10}\right)^2
\]
Thus,
\[
g_h
=
g\left(\frac{10}{11}\right)^2
=
\frac{100}{121}g
\]
Numerically,
\[
\frac{100}{121}\approx0.826g
\]
\(\textbf{Small-altitude approximation:}\)
For \(h\ll R\),
\[
g_h\approx g\left(1-\frac{2h}{R}\right)
\]
Substituting \(h=R/10\),
\[
g_h\approx g\left(1-\frac{2}{10}\right)=0.80g
\]
The approximate expression is obtained by truncating higher-order terms and must not be treated as identical to the exact value.
\(\textbf{Final answer:}\)
\[
g_h^{\text{exact}}=\frac{100}{121}g,\qquad
g_h^{\text{approx}}=0.80g
\]
111. A planet has radius \(6400\,\text{km}\) and surface acceleration due to gravity \(9.8\,\text{m s}^{-2}\). Using the small-altitude approximation, what are \(g_h\) and the weight of a \(5.0\,\text{kg}\) body at altitude \(32\,\text{km}\)?
ⓐ. \(g_h=9.60\,\text{m s}^{-2},\quad W_h=48.0\,\text{N}\)
ⓑ. \(g_h=9.702\,\text{m s}^{-2},\quad W_h=48.51\,\text{N}\)
ⓒ. \(g_h=9.751\,\text{m s}^{-2},\quad W_h=48.76\,\text{N}\)
ⓓ. \(g_h=9.80\,\text{m s}^{-2},\quad W_h=49.0\,\text{N}\)
Correct Answer: \(g_h=9.702\,\text{m s}^{-2},\quad W_h=48.51\,\text{N}\)
Explanation: \(\textbf{Check the small parameter:}\)
\[
\frac{h}{R}
=
\frac{32}{6400}
=0.005
\]
Since this ratio is much smaller than \(1\), the small-altitude approximation is appropriate.
Use
\[
g_h\approx g\left(1-\frac{2h}{R}\right)
\]
Therefore,
\[
g_h
\approx
9.8\left(1-2\times0.005\right)
\]
Thus,
\[
g_h=9.8(0.99)
\]
Hence,
\[
g_h=9.702\,\text{m s}^{-2}
\]
\(\textbf{Weight at that altitude:}\)
\[
W_h=mg_h
\]
Substituting,
\[
W_h=(5.0)(9.702)=48.51\,\text{N}
\]
The approximate fractional decrease in \(g\) is \(2h/R=0.01\), or \(1\%\).
\(\textbf{Final answer:}\)
\[
\boxed{g_h\approx9.702\,\text{m s}^{-2},\qquad
W_h\approx48.51\,\text{N}}
\]
112. At altitude \(h=R/20\) above a spherical planet, one student uses the exact expression for \(g_h\), while another uses the first-order small-altitude approximation. Which statement is correct?
ⓐ. Exact: \(0.90g\); approximation: \(\dfrac{400}{441}g\), so the approximation overestimates
ⓑ. Exact: \(\dfrac{400}{441}g\); approximation: \(\dfrac{400}{441}g\), so they are identical
ⓒ. Exact: \(\dfrac{400}{441}g\); approximation: \(0.90g\), with an error greater than \(5\%\)
ⓓ. Exact: \(\dfrac{400}{441}g\); approximation: \(0.90g\), with an underestimate below \(1\%\)
Correct Answer: Exact: \(\dfrac{400}{441}g\); approximation: \(0.90g\), with an underestimate below \(1\%\)
Explanation: \(\textbf{Exact value:}\)
For
\[
h=\frac R{20},
\]
\[
\frac{g_h}{g}
=
\left(\frac{R}{R+R/20}\right)^2
\]
Thus,
\[
\frac{g_h}{g}
=
\left(\frac{20}{21}\right)^2
=
\frac{400}{441}
\]
Numerically,
\[
\frac{400}{441}\approx0.9070
\]
\(\textbf{First-order approximation:}\)
\[
\frac{g_h}{g}
\approx
1-\frac{2h}{R}
\]
Hence,
\[
\frac{g_h}{g}
\approx1-\frac{2}{20}=0.90
\]
The approximation is therefore slightly smaller than the exact result.
The fractional error relative to the exact result is approximately
\[
\frac{0.9070-0.9000}{0.9070}\times100\%\approx0.8\%
\]
Thus the error is below \(1\%\), illustrating why the approximation is useful when \(h/R\) is sufficiently small.
\(\textbf{Final answer:}\) The exact ratio is \(400/441\), while the first-order approximation gives \(0.90\) and slightly underestimates it.
113. The same physical altitude \(h=100\,\text{km}\) is considered above two spherical planets. Planet A has radius \(5000\,\text{km}\), while planet B has radius \(500\,\text{km}\). A student proposes using
\[
g_h\approx g\left(1-\frac{2h}{R}\right)
\]
for both planets. Which assessment is correct?
ⓐ. The approximation is equally reliable because the numerical altitude \(h\) is the same
ⓑ. The approximation is more reliable for B because planet B has the smaller radius
ⓒ. The approximation is much more reliable for A because the fractional altitude is much smaller for A than for B
ⓓ. The approximation is exact for both planets because \(h\lt R\) in each case
Correct Answer: The approximation is much more reliable for A because the fractional altitude is much smaller for A than for B
Explanation: \(\textbf{The validity criterion is dimensionless:}\)
The approximation requires
\[
\frac{h}{R}\ll1.
\]
For planet A,
\[
\frac{h}{R_A}
=
\frac{100}{5000}=0.02
\]
Its approximate gravity ratio is
\[
\frac{g_h}{g}
\approx1-2(0.02)=0.96
\]
For planet B,
\[
\frac{h}{R_B}
=
\frac{100}{500}=0.20
\]
This ratio is not very small, so the first-order approximation is much less reliable.
It predicts
\[
\frac{g_h}{g}\approx1-2(0.20)=0.60
\]
\(\textbf{Check planet B with the exact relation:}\)
\[
\frac{g_h}{g}
=
\left(\frac{R_B}{R_B+h}\right)^2
\]
Therefore,
\[
\frac{g_h}{g}
=
\left(\frac{500}{600}\right)^2
=
\frac{25}{36}
\approx0.694
\]
The sizeable difference between \(0.60\) and \(0.694\) shows that numerical altitude alone does not determine approximation quality; the controlling parameter is \(h/R\).
\(\textbf{Final answer:}\) The approximation is substantially more reliable for planet A than for planet B.
114. For a spherical planet of radius \(R\) and surface acceleration due to gravity \(g\), four measurements are reported:
| Measurement | Altitude | Reported \(g_h\) |
|---|
| P | \(R/2\) | \(4g/9\) |
| Q | \(R\) | \(g/4\) |
| R | \(2R\) | \(g/9\) |
| S | \(3R\) | \(g/12\) |
Which measurement is inconsistent with the exact altitude relation, and what should its reported value be?
ⓐ. P; it should be \(g/2\)
ⓑ. Q; it should be \(g/3\)
ⓒ. R; it should be \(g/8\)
ⓓ. S; it should be \(g/16\)
Correct Answer: S; it should be \(g/16\)
Explanation: The exact relation outside a spherical planet is
\[
g_h=g\left(\frac{R}{R+h}\right)^2
\]
For P,
\[
h=\frac R2
\]
gives
\[
\frac{g_h}{g}
=\left(\frac{R}{3R/2}\right)^2
=\frac49,
\]
so P is consistent.
For Q,
\[
h=R\Rightarrow \frac{g_h}{g}=\left(\frac12\right)^2=\frac14
\]
For R,
\[
h=2R\Rightarrow \frac{g_h}{g}=\left(\frac13\right)^2=\frac19
\]
For S,
\[
h=3R\Rightarrow R+h=4R
\]
Hence,
\[
\frac{g_h}{g}=\left(\frac14\right)^2=\frac1{16}
\]
Thus only measurement S is inconsistent.
\(\textbf{Final answer:}\)
\[
g_{3R}=\frac{g}{16}
\]
115. For a spherical planet, let
\[
y=\frac{1}{\sqrt{g_h}},
\]
where \(g_h\) is the acceleration due to gravity at altitude \(h\).
A graph of \(y\) against \(h\) is found to be a straight line with vertical-axis intercept \(b\) and slope \(s\).
Which pair correctly gives the planet's radius \(R\) and mass \(M\)?
ⓐ. \(R=bs,\quad M=\dfrac{s^2}{G}\)
ⓑ. \(R=\dfrac{b}{s},\quad M=\dfrac{1}{Gs^2}\)
ⓒ. \(R=\dfrac{s}{b},\quad M=\dfrac{1}{Gb^2}\)
ⓓ. \(R=\dfrac{b}{s},\quad M=\dfrac{s^2}{G}\)
Correct Answer: \(R=\dfrac{b}{s},\quad M=\dfrac{1}{Gs^2}\)
Explanation: \(\textbf{Transform the inverse-square relation:}\)
At altitude \(h\),
\[
g_h=\frac{GM}{(R+h)^2}
\]
Taking the reciprocal square root,
\[
\frac{1}{\sqrt{g_h}}
=
\frac{R+h}{\sqrt{GM}}
\]
Therefore,
\[
y=
\frac{R}{\sqrt{GM}}
+
\frac{1}{\sqrt{GM}}h
\]
This has the straight-line form
\[
y=b+sh
\]
Hence the slope is
\[
s=\frac{1}{\sqrt{GM}},
\]
while the intercept is
\[
b=\frac{R}{\sqrt{GM}}
\]
Dividing intercept by slope,
\[
\frac{b}{s}=R
\]
Also,
\[
s^2=\frac{1}{GM},
\]
so
\[
M=\frac{1}{Gs^2}
\]
This transformed graph therefore allows both planetary radius and mass to be inferred.
\(\textbf{Final answer:}\)
\[
\boxed{R=\frac{b}{s},\qquad M=\frac{1}{Gs^2}}
\]
116. A spherical planet has radius \(6.0\times10^6\,\text{m}\). At an altitude \(3.0\times10^6\,\text{m}\), the acceleration due to gravity is \(4.0\,\text{m s}^{-2}\). Take \(G=6.67\times10^{-11}\,\text{N m}^2\text{kg}^{-2}\). Assuming the planet to be a uniform sphere, its mass and mean density are approximately
ⓐ. \(2.43\times10^{24}\,\text{kg}\) and \(2.69\times10^3\,\text{kg m}^{-3}\)
ⓑ. \(4.86\times10^{24}\,\text{kg}\) and \(2.69\times10^3\,\text{kg m}^{-3}\)
ⓒ. \(4.86\times10^{24}\,\text{kg}\) and \(5.37\times10^3\,\text{kg m}^{-3}\)
ⓓ. \(9.72\times10^{24}\,\text{kg}\) and \(5.37\times10^3\,\text{kg m}^{-3}\)
Correct Answer: \(4.86\times10^{24}\,\text{kg}\) and \(5.37\times10^3\,\text{kg m}^{-3}\)
Explanation: \(\textbf{Recover the surface gravity:}\)
Here,
\[
h=\frac R2
\]
Therefore,
\[
g_h=g\left(\frac{R}{R+h}\right)^2
=g\left(\frac{2}{3}\right)^2
\]
Thus,
\[
4.0=\frac49g,
\]
which gives
\[
g=9.0\,\text{m s}^{-2}
\]
\(\textbf{Determine the planetary mass:}\)
Using
\[
g=\frac{GM}{R^2},
\]
\[
M=\frac{gR^2}{G}
\]
Substitution gives
\[
M=
\frac{(9.0)(6.0\times10^6)^2}
{6.67\times10^{-11}}
\]
Hence,
\[
M\approx4.86\times10^{24}\,\text{kg}
\]
\(\textbf{Determine the mean density:}\)
\[
\rho=\frac{M}{(4/3)\pi R^3}
\]
Therefore,
\[
\rho
=
\frac{4.86\times10^{24}}
{(4/3)\pi(6.0\times10^6)^3}
\]
This gives approximately
\[
\rho\approx5.37\times10^3\,\text{kg m}^{-3}
\]
\(\textbf{Final answer:}\)
\[
M\approx4.86\times10^{24}\,\text{kg},\qquad
\rho\approx5.37\times10^3\,\text{kg m}^{-3}
\]
117. Uniform spherical planet A has radius \(R\) and mean density \(\rho\). Uniform spherical planet B has mean density \(3\rho\). On each planet, \(g_h\) is measured at an altitude equal to that planet's own radius. The measured value for B is six times that for A. What are \(R_B/R\) and \(M_B/M_A\)?
ⓐ. \(3,\ 27\)
ⓑ. \(2,\ 12\)
ⓒ. \(2,\ 24\)
ⓓ. \(\sqrt2,\ 6\sqrt2\)
Correct Answer: \(2,\ 24\)
Explanation: \(\textbf{Gravity at an altitude equal to the planetary radius:}\)
For any planet,
\[
h=R_p
\]
gives
\[
g_h=g_s\left(\frac{R_p}{2R_p}\right)^2
=\frac{g_s}{4}
\]
For a uniform sphere,
\[
g_s=\frac43\pi G\rho R_p
\]
Therefore at \(h=R_p\),
\[
g_h=\frac13\pi G\rho R_p
\]
Thus the ratio for the two planets is
\[
\frac{g_{h,B}}{g_{h,A}}
=
\frac{\rho_B R_B}{\rho R}
\]
The measured ratio is \(6\), while
\[
\rho_B=3\rho
\]
Hence,
\[
6=3\frac{R_B}{R}
\]
Therefore,
\[
\frac{R_B}{R}=2
\]
\(\textbf{Infer the mass ratio:}\)
For uniform spheres,
\[
M\propto\rho R^3
\]
Thus,
\[
\frac{M_B}{M_A}
=
3(2)^3
=24
\]
\(\textbf{Final answer:}\)
\[
\boxed{\frac{R_B}{R}=2,\qquad
\frac{M_B}{M_A}=24}
\]
118. A spherical planet of mass \(M\) and radius \(R\) expands without loss of mass until its radius becomes \(2R\). A test body is kept fixed in space at a point that was originally at altitude \(R\) above the planet's surface. If the original surface gravity was \(g\), which statement is correct immediately after the expansion?
ⓐ. The body's weight doubles, and the new surface gravity is \(g/2\)
ⓑ. The body's weight is unchanged, and the new surface gravity is \(g/4\)
ⓒ. The body's weight becomes one-fourth, and the new surface gravity remains \(g\)
ⓓ. The body's weight is unchanged, and the new surface gravity remains \(g\)
Correct Answer: The body's weight is unchanged, and the new surface gravity is \(g/4\)
Explanation: Before the expansion, the test body is at altitude \(R\).
Its centre-distance is therefore
\[
r=R+R=2R
\]
Its gravitational acceleration is
\[
g_{\text{body}}
=
\frac{GM}{(2R)^2}
=\frac{GM}{4R^2}
\]
After expansion, the planet's new radius is \(2R\).
The test body has not moved in space, so its centre-distance remains exactly \(2R\).
It is now located at the new surface.
Because both the source mass and centre-distance are unchanged, the gravitational acceleration at the body's position is unchanged.
Therefore its weight is also unchanged.
The original surface gravity was
\[
g=\frac{GM}{R^2}
\]
The new surface gravity is
\[
g'=\frac{GM}{(2R)^2}
=\frac g4
\]
The problem illustrates why altitude and centre-distance must not be confused.
\(\textbf{Final answer:}\) The weight is unchanged and the new surface gravity is \(g/4\).
119. The same small physical altitude \(h\) is considered above two uniform spherical planets A and B of equal mean density. Using the small-altitude approximation, the fractional decreases in \(g\) are \(2\%\) on A and \(5\%\) on B. What are \(R_A/R_B\) and \(g_A/g_B\), where \(g_A\) and \(g_B\) denote their surface gravities?
ⓐ. \(\dfrac25,\ \dfrac25\)
ⓑ. \(\dfrac52,\ 1\)
ⓒ. \(\dfrac52,\ \dfrac52\)
ⓓ. \(\dfrac54,\ \dfrac52\)
Correct Answer: \(\dfrac52,\ \dfrac52\)
Explanation: \(\textbf{Use the approximate fractional decrease:}\)
For \(h\ll R\),
\[
\frac{g-g_h}{g}\approx\frac{2h}{R}
\]
For planet A,
\[
0.02=\frac{2h}{R_A}
\]
For planet B,
\[
0.05=\frac{2h}{R_B}
\]
Dividing the two relations,
\[
\frac{0.02}{0.05}
=
\frac{R_B}{R_A}
\]
Hence,
\[
\frac{R_A}{R_B}=\frac{0.05}{0.02}=\frac52
\]
\(\textbf{Use the equal-density condition:}\)
For uniform spherical planets,
\[
g=\frac43\pi G\rho R
\]
At equal mean density,
\[
g\propto R
\]
Therefore,
\[
\frac{g_A}{g_B}
=
\frac{R_A}{R_B}
=\frac52
\]
The smaller fractional decrease occurs on the planet with the larger radius because the same physical altitude represents a smaller value of \(h/R\).
\(\textbf{Final answer:}\)
\[
\boxed{\frac{R_A}{R_B}=\frac52,\qquad
\frac{g_A}{g_B}=\frac52}
\]
120. Consider the following statements about variation of \(g\) with altitude.
Statement I: The small-altitude approximation requires \(h/R\ll1\).
Statement II: In that approximation, the fractional decrease in \(g\) is approximately \(2h/R\).
Statement III: If two planets have the same value of \(h/R\), then the exact ratio \(g_h/g\) is the same for both.
Statement IV: The first-order approximation may replace the exact expression even when \(h\) is comparable with \(R\), because both predict a decrease in \(g\).
Statement V: A decrease of local \(g\) with altitude implies that the universal gravitational constant \(G\) decreases with altitude.
Which statements are correct?
ⓐ. I, II and III only
ⓑ. I, III and IV only
ⓒ. II, IV and V only
ⓓ. I, II, III and V only
Correct Answer: I, II and III only
Explanation: Statement I is correct because the approximation is obtained by expanding in the small dimensionless parameter \(h/R\).
Statement II is also correct since
\[
g_h\approx g\left(1-\frac{2h}{R}\right).
\]
Thus the approximate fractional decrease is \(2h/R\).
Statement III follows from the exact expression
\[
\frac{g_h}{g}
=
\frac{1}{(1+h/R)^2}.
\]
Only the ratio \(h/R\) appears, so equal values of \(h/R\) give equal normalized gravity values.
Statement IV is incorrect because the neglected higher-order terms become important when \(h\) is not small compared with \(R\).
Statement V is also incorrect because \(G\) is universal.
The local decrease occurs because the centre-distance increases from \(R\) to \(R+h\).
Thus only Statements I, II and III are correct.
