101. At a height \(h\) above Earth’s surface, the distance from Earth’s centre to a body is
ⓐ. \(h\)
ⓑ. \(R_E+h\)
ⓒ. \(R_E-h\)
ⓓ. \(R_E\)
Correct Answer: \(R_E+h\)
Explanation: The height \(h\) is measured from Earth’s surface, not from Earth’s centre. In gravitational formulas for Earth, the relevant distance is the centre-to-body distance. A body on the surface is already at a distance \(R_E\) from Earth’s centre. Raising it by \(h\) makes the distance \(R_E+h\). This centre-distance idea is essential because gravitational acceleration varies with the square of this distance.
102. The acceleration due to gravity at height \(h\) above Earth’s surface is given by
ⓐ. \(g_h=g\left(\frac{R_E+h}{R_E}\right)^2\)
ⓑ. \(g_h=g\left(\frac{R_E}{R_E+h}\right)^2\)
ⓒ. \(g_h=g\left(1+\frac{h}{R_E}\right)\)
ⓓ. \(g_h=g\left(\frac{h}{R_E}\right)^2\)
Correct Answer: \(g_h=g\left(\frac{R_E}{R_E+h}\right)^2\)
Explanation: At Earth’s surface, \(g=G\frac{M_E}{R_E^2}\). At height \(h\), the centre-to-body distance becomes \(R_E+h\), so \(g_h=G\frac{M_E}{(R_E+h)^2}\). Dividing the height expression by the surface expression gives \(g_h=g\left(\frac{R_E}{R_E+h}\right)^2\). Since \(R_E+h\) is larger than \(R_E\), the fraction is less than \(1\). This shows that \(g\) decreases with altitude.
103. A satellite is at a height equal to Earth’s radius above the surface. If surface gravity is \(g\), the value of \(g\) at the satellite is
ⓐ. \(\frac{g}{4}\)
ⓑ. \(\frac{g}{2}\)
ⓒ. \(4g\)
ⓓ. \(2g\)
Correct Answer: \(\frac{g}{4}\)
Explanation: \( \textbf{Given condition:} \) the height is \(h=R_E\).
The distance from Earth’s centre is
\[
r=R_E+h=R_E+R_E=2R_E
\]
Use the altitude relation:
\[
g_h=g\left(\frac{R_E}{R_E+h}\right)^2
\]
Substitute \(h=R_E\):
\[
g_h=g\left(\frac{R_E}{2R_E}\right)^2
\]
\[
g_h=g\left(\frac{1}{2}\right)^2
\]
\[
g_h=\frac{g}{4}
\]
\( \textbf{Final answer:} \) \(\frac{g}{4}\), because the distance from Earth’s centre doubles and the inverse-square factor becomes one-fourth.
104. A notebook line says, “At height \(h\), use \(g_h=G\frac{M_E}{h^2}\).” The main error in this line is that it uses
ⓐ. body mass \(m\) instead of Earth mass \(M_E\)
ⓑ. linear \(h\) dependence instead of inverse-square
ⓒ. height \(h\) instead of centre-distance \(R_E+h\)
ⓓ. \(G\) instead of surface value \(g\)
Correct Answer: height \(h\) instead of centre-distance \(R_E+h\)
Explanation: The inverse-square formula for Earth outside its surface uses the distance from Earth’s centre. At a height \(h\), that distance is \(R_E+h\), not \(h\). The suitable expression is \(g_h=G\frac{M_E}{(R_E+h)^2}\). Earth’s mass \(M_E\) is correctly present because Earth is the source of the gravitational field. The inverse-square dependence is also correct; the mistake is the distance chosen for the denominator.
105. For \(h\ll R_E\), the approximate expression for gravity at height \(h\) is
ⓐ. \(g_h\approx g\left(1-\frac{h^2}{R_E^2}\right)\)
ⓑ. \(g_h\approx g\left(1-\frac{2h}{R_E}\right)\)
ⓒ. \(g_h\approx g\left(\frac{2h}{R_E}\right)\)
ⓓ. \(g_h\approx g\left(1+\frac{2h}{R_E}\right)\)
Correct Answer: \(g_h\approx g\left(1-\frac{2h}{R_E}\right)\)
Explanation: Start with the exact altitude relation \(g_h=g\left(\frac{R_E}{R_E+h}\right)^2\). This can be written as \(g_h=g\left(1+\frac{h}{R_E}\right)^{-2}\). For a small value of \(\frac{h}{R_E}\), the binomial approximation \((1+x)^{-2}\approx1-2x\) may be used. Substituting \(x=\frac{h}{R_E}\) gives \(g_h\approx g\left(1-\frac{2h}{R_E}\right)\). The approximation is meant for small heights compared with Earth’s radius, not for very large altitudes.
106. At a height \(h\) above Earth, the value of \(g_h\) is \(0.81g\). The ratio \(\frac{R_E+h}{R_E}\) is
ⓐ. \(\frac{9}{10}\)
ⓑ. \(\frac{81}{100}\)
ⓒ. \(\frac{100}{81}\)
ⓓ. \(\frac{10}{9}\)
Correct Answer: \(\frac{10}{9}\)
Explanation: \( \textbf{Use the altitude relation:} \)
\[
g_h=g\left(\frac{R_E}{R_E+h}\right)^2
\]
Given:
\[
g_h=0.81g=\frac{81}{100}g
\]
So,
\[
\frac{81}{100}=\left(\frac{R_E}{R_E+h}\right)^2
\]
Taking square root:
\[
\frac{9}{10}=\frac{R_E}{R_E+h}
\]
Invert both sides:
\[
\frac{R_E+h}{R_E}=\frac{10}{9}
\]
\( \textbf{Final answer:} \) \(\frac{10}{9}\), with the square root step needed because \(g\) depends on the square of the centre-distance ratio.
107. A graph of \(g_h\) against height \(h\) above Earth’s surface should show that \(g_h\)
ⓐ. decreases as \(R_E+h\) increases
ⓑ. remains constant as \(R_E+h\) increases
ⓒ. increases as \(R_E+h\) increases
ⓓ. becomes zero when \(h\) first exceeds \(0\)
Correct Answer: decreases as \(R_E+h\) increases
Explanation: Above Earth’s surface, \(g_h=G\frac{M_E}{(R_E+h)^2}\). As \(h\) increases, the distance \(R_E+h\) from Earth’s centre increases. Since this distance is squared in the denominator, \(g_h\) decreases with height. The decrease is not linear for large heights; it follows an inverse-square pattern with centre-distance. Gravity is weaker at high altitude, but it does not suddenly become zero just above the surface.
108. The table gives values of height for the same Earth.
| Case | Height above surface | Centre-distance |
|---|
| P | \(0\) | \(R_E\) |
| Q | \(R_E\) | \(2R_E\) |
| R | \(2R_E\) | \(3R_E\) |
The corresponding values of \(g_h\) are
ⓐ. \(g,4g,9g\)
ⓑ. \(g,2g,3g\)
ⓒ. \(g,\frac{g}{2},\frac{g}{3}\)
ⓓ. \(g,\frac{g}{4},\frac{g}{9}\)
Correct Answer: \(g,\frac{g}{4},\frac{g}{9}\)
Explanation: At the surface, the centre-distance is \(R_E\), so the value is \(g\). In case Q, the centre-distance is \(2R_E\), so the inverse-square factor is \(\left(\frac{R_E}{2R_E}\right)^2=\frac{1}{4}\). In case R, the centre-distance is \(3R_E\), so the factor is \(\left(\frac{R_E}{3R_E}\right)^2=\frac{1}{9}\). The height itself is not the denominator; the full centre-distance is used. The values therefore follow \(g,\frac{g}{4},\frac{g}{9}\).
109. Inside Earth, using the uniform-density model, the value of \(g\) at depth \(d\) below the surface is
ⓐ. \(g_d=g\left(\frac{d}{R_E}\right)^2\)
ⓑ. \(g_d=g\left(1+\frac{d}{R_E}\right)^2\)
ⓒ. \(g_d=g\left(1-\frac{d}{R_E}\right)\)
ⓓ. \(g_d=g\left(\frac{R_E}{R_E-d}\right)^2\)
Correct Answer: \(g_d=g\left(1-\frac{d}{R_E}\right)\)
Explanation: In the uniform-density model, the gravitational acceleration inside Earth decreases linearly with depth. At depth \(d\), the distance from Earth’s centre is \(R_E-d\). Since the interior field is proportional to the distance from the centre, \(g_d=g\frac{R_E-d}{R_E}\). This becomes \(g_d=g\left(1-\frac{d}{R_E}\right)\). The inverse-square formula is not used inside Earth in this uniform solid-sphere model.
110. At the centre of Earth, the value of \(g\) in the uniform-density model is
ⓐ. \(2g\)
ⓑ. \(0\)
ⓒ. \(g\)
ⓓ. \(\frac{g}{2}\)
Correct Answer: \(0\)
Explanation: At Earth’s centre, gravitational pulls from different directions cancel by symmetry. The depth is \(d=R_E\), so the depth formula gives \(g_d=g\left(1-\frac{R_E}{R_E}\right)=0\). This does not mean Earth has no mass at the centre; it means the net gravitational field there is zero. The mass outside any chosen central point pulls equally in all directions. The zero value comes from vector cancellation, not from absence of gravitational interaction.
111. A body is taken to a depth \(\frac{R_E}{4}\) below Earth’s surface. In the uniform-density model, the acceleration due to gravity there is
ⓐ. \(\frac{g}{2}\)
ⓑ. \(\frac{5g}{4}\)
ⓒ. \(\frac{g}{4}\)
ⓓ. \(\frac{3g}{4}\)
Correct Answer: \(\frac{3g}{4}\)
Explanation: \( \textbf{Depth relation:} \)
\[
g_d=g\left(1-\frac{d}{R_E}\right)
\]
Given:
\[
d=\frac{R_E}{4}
\]
Substitute:
\[
g_d=g\left(1-\frac{R_E/4}{R_E}\right)
\]
\[
g_d=g\left(1-\frac{1}{4}\right)
\]
\[
g_d=\frac{3g}{4}
\]
\( \textbf{Final answer:} \) \(\frac{3g}{4}\), because depth variation is linear in this model, not inverse-square.
112. Two locations are compared: Point P is at height \(h\) above Earth’s surface, and Point Q is at depth \(h\) below Earth’s surface, where \(h\ll R_E\). Using standard approximations, the decrease in \(g\) is
ⓐ. nearly twice as large at P as at Q
ⓑ. the same because height and depth enter identically
ⓒ. nearly twice as large at Q as at P
ⓓ. exactly zero at both points
Correct Answer: nearly twice as large at P as at Q
Explanation: For a small height \(h\), the approximate altitude result is \(g_h\approx g\left(1-\frac{2h}{R_E}\right)\). The decrease is therefore about \(\frac{2gh}{R_E}\). For a depth \(h\), the uniform-density result is \(g_d=g\left(1-\frac{h}{R_E}\right)\). The decrease is \(\frac{gh}{R_E}\). For the same small value of \(h\), the altitude decrease is approximately twice the depth decrease.
113. A graph of \(g\) against depth \(d\) below Earth’s surface, assuming uniform density, is expected to be
ⓐ. an inverse-square curve increasing toward the centre
ⓑ. a horizontal line equal to \(g\) at all depths
ⓒ. a straight line decreasing from \(g\) at \(d=0\) to \(0\) at \(d=R_E\)
ⓓ. a straight line increasing from \(0\) at \(d=0\) to \(g\) at \(d=R_E\)
Correct Answer: a straight line decreasing from \(g\) at \(d=0\) to \(0\) at \(d=R_E\)
Explanation: The depth formula in the uniform-density model is \(g_d=g\left(1-\frac{d}{R_E}\right)\). This is a linear relation in \(d\). At the surface, \(d=0\), so \(g_d=g\). At the centre, \(d=R_E\), so \(g_d=0\). The graph is therefore a straight decreasing line from surface to centre.
114. The statement that best contrasts altitude and depth variation of \(g\) is
ⓐ. altitude uses \(R_E+h\), while depth uses the uniform-Earth linear law
ⓑ. altitude and depth both use inverse-square distance measured from the surface
ⓒ. \(g\) increases with altitude but decreases with depth
ⓓ. \(g\) is zero at height \(R_E\) and depth \(R_E\)
Correct Answer: altitude uses \(R_E+h\), while depth uses the uniform-Earth linear law
Explanation: Above Earth’s surface, the point lies outside the spherical Earth, so \(g_h=G\frac{M_E}{(R_E+h)^2}\). Inside a uniform Earth, only the enclosed mass effectively contributes, giving \(g_d=g\left(1-\frac{d}{R_E}\right)\). These are different physical situations and require different formulas. Gravity at height \(R_E\) is \(\frac{g}{4}\), not zero. The depth result becomes zero only at the centre in the uniform-density model.
115. Earth’s rotation affects the apparent value of \(g\) because a body on Earth also needs
ⓐ. escape-speed requirement from Earth’s rotation
ⓑ. a lunar outward force caused by Earth’s rotation
ⓒ. zero gravitational pull at the rotating equator
ⓓ. centripetal acceleration from Earth’s rotation
Correct Answer: centripetal acceleration from Earth’s rotation
Explanation: A body on Earth’s rotating surface moves in a circular path around Earth’s rotation axis. Part of the gravitational pull is associated with providing the required centripetal acceleration. The measured or effective value of \(g\) is therefore slightly reduced compared with the purely gravitational value. This reduction is largest at the equator and zero at the poles. The effect is not because gravity disappears at the equator; it is a rotation-related correction.
116. The reduction in effective \(g\) due to Earth’s rotation is maximum at the equator because there
ⓐ. the distance from Earth’s rotation axis is zero
ⓑ. Earth’s mass becomes minimum
ⓒ. the universal gravitational constant becomes smaller
ⓓ. the distance from Earth’s rotation axis is maximum
Correct Answer: the distance from Earth’s rotation axis is maximum
Explanation: The rotational effect depends on the distance from Earth’s rotation axis. At the equator, a point is farthest from the rotation axis, so the required centripetal acceleration is largest. This produces the greatest reduction in effective \(g\). At the poles, the distance from the rotation axis is zero, so this rotational reduction vanishes. The value of \(G\) does not change from equator to pole.
117. At latitude \(\lambda\), the standard formula for effective acceleration due to Earth’s rotation is
ⓐ. \(g'=g-\omega^2R_E\cos^2\lambda\)
ⓑ. \(g'=\omega^2R_E\cos^2\lambda-g\)
ⓒ. \(g'=g+\omega^2R_E\cos^2\lambda\)
ⓓ. \(g'=g-\omega^2R_E\sin^2\lambda\)
Correct Answer: \(g'=g-\omega^2R_E\cos^2\lambda\)
Explanation: Earth’s rotation reduces the effective value of \(g\) because the rotational effect acts opposite to the apparent weight contribution. The latitude dependence is represented by the factor \(\cos^2\lambda\). At the equator, \(\lambda=0^\circ\), so \(\cos^2\lambda=1\) and the reduction is maximum. At the poles, \(\lambda=90^\circ\), so \(\cos^2\lambda=0\) and the rotational reduction is zero. The minus sign shows reduction, not reversal of gravity.
118. If only Earth’s shape is considered, \(g\) is smaller at the equator than at the poles because
ⓐ. the equatorial radius is larger than the polar radius
ⓑ. Earth’s mass is absent at the equator
ⓒ. \(G\) is larger at the poles than at the equator
ⓓ. the polar radius is larger than the equatorial radius
Correct Answer: the equatorial radius is larger than the polar radius
Explanation: Earth is slightly oblate, so its equatorial radius is larger than its polar radius. Since gravitational acceleration near the surface varies approximately as \(g=\frac{GM_E}{R^2}\), a larger radius gives a smaller value of \(g\). Therefore the radius effect makes \(g\) smaller at the equator and larger at the poles. This shape-based reason is separate from the rotational reduction. Both effects act to make the observed \(g\) smaller at the equator.
119. The following table compares two causes of latitude variation in \(g\).
| Cause | Effect at equator | Effect at poles |
|---|
| P. Rotation | Maximum reduction in effective \(g\) | No rotational reduction |
| Q. Oblate shape | Larger radius gives smaller \(g\) | Smaller radius gives larger \(g\) |
The table is
ⓐ. Q is correct and P is wrong
ⓑ. P and Q are both wrong
ⓒ. P and Q are both correct
ⓓ. P is correct and Q is wrong
Correct Answer: P and Q are both correct
Explanation: Rotation reduces the effective value of \(g\), and the reduction is greatest at the equator because the distance from the rotation axis is largest there. At the poles, this rotational reduction is zero. Earth’s oblate shape also matters because the equatorial radius is larger than the polar radius. A larger radius gives a smaller gravitational acceleration for the same central mass. Thus both listed effects support a smaller effective \(g\) at the equator than at the poles.
120. An object is moved from the equator to the pole. Ignoring local geological variations, its mass and weight change as
ⓐ. mass remains the same and weight increases
ⓑ. mass decreases and weight increases
ⓒ. mass increases and weight decreases
ⓓ. both mass and weight remain exactly unchanged
Correct Answer: mass remains the same and weight increases
Explanation: The mass of the object is not changed by moving it from one latitude to another. Weight depends on the effective value of \(g\), through \(W=mg\). At the pole, the rotational reduction is absent and the polar radius is smaller than the equatorial radius. Both reasons make the effective value of \(g\) greater at the pole than at the equator. The object therefore has the same mass but a slightly larger weight at the pole.
