Gravitation — Practice Questions with Answers

Newton's law of universal gravitation states that every particle in the universe attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them, acting along the line joining them.

$F = \dfrac{Gm_1m_2}{r^2}$, where $G$ is the universal gravitational constant ($\approx 6.674\times10^{-11}\text{ N·m}^2/\text{kg}^2$).

Consider a body of mass $m$ on Earth's surface (mass $M$, radius $R$). The gravitational force on it equals its weight:

$mg = \dfrac{GMm}{R^2}$

The mass $m$ cancels: $g = \dfrac{GM}{R^2}$

This shows $g$ depends on Earth's mass and radius, but not on the mass of the falling body — explaining why all objects fall with the same acceleration in the absence of air resistance.

At height $h$ above the surface, distance from Earth's centre becomes $(R+h)$, so: $g_h = \dfrac{GM}{(R+h)^2}$

Comparing with surface gravity $g = GM/R^2$: $\dfrac{g_h}{g} = \dfrac{R^2}{(R+h)^2}$

For $h \ll R$, using the binomial approximation: $g_h \approx g\left(1 - \dfrac{2h}{R}\right)$ — showing $g$ decreases approximately linearly with altitude for small heights.

At depth $d$ below the surface, only the mass enclosed within radius $(R-d)$ contributes to gravitational pull (assuming uniform density) — the shell of material outside this radius exerts no net gravitational force at that interior point.

This gives $g_d = g\left(1 - \dfrac{d}{R}\right)$ — showing $g$ decreases linearly with depth, becoming zero exactly at the Earth's centre ($d=R$).

Escape velocity is the minimum speed needed for an object to escape a planet's gravitational field permanently, reaching infinity with zero final kinetic energy.

Using energy conservation: initial KE + initial PE = final KE + final PE (final KE = 0, final PE = 0 at infinity):

$\dfrac{1}{2}mv_e^2 + \left(-\dfrac{GMm}{R}\right) = 0$

$v_e = \sqrt{\dfrac{2GM}{R}}$, which can also be written as $v_e = \sqrt{2gR}$.

First law (law of orbits): Each planet moves in an elliptical orbit with the Sun at one focus.

Second law (law of areas): The line joining a planet to the Sun sweeps out equal areas in equal intervals of time.

Third law (law of periods): The square of a planet's orbital period is directly proportional to the cube of the semi-major axis of its orbit: $T^2 \propto a^3$.

For a satellite in a circular orbit, gravitational force provides the necessary centripetal force:

$\dfrac{GMm}{r^2} = \dfrac{mv_o^2}{r}$

Solving: $v_o = \sqrt{\dfrac{GM}{r}}$

For an orbit close to Earth's surface, $r \approx R$: $v_o = \sqrt{gR} \approx 7.9\text{ km/s}$.

Gravitational potential energy is the energy possessed by a body due to its position within a gravitational field — the work done in bringing it from a reference point (usually infinity) to its current position.

Since gravity is always attractive, work must be done against gravity to move a body to infinity (where PE is defined as zero) — meaning gravity does positive work as a body moves inward. This makes the PE at any finite distance less than the zero reference at infinity, hence negative: $U = -\dfrac{GMm}{r}$.

A geostationary satellite is one that appears stationary when observed from a fixed point on Earth's surface, since its orbital period exactly matches Earth's rotational period.

Essential conditions: (1) It must orbit in the equatorial plane (same plane as Earth's rotation). (2) It must have an orbital period of exactly 24 hours (matching Earth's rotation), and must orbit in the same direction as Earth's rotation (west to east), at a specific altitude (~36,000 km) satisfying this period requirement.

For a satellite/planet in circular orbit, gravitational force provides centripetal force:

$\dfrac{GMm}{r^2} = m\omega^2r$, where $\omega = \dfrac{2\pi}{T}$

$\dfrac{GM}{r^2} = \left(\dfrac{2\pi}{T}\right)^2r = \dfrac{4\pi^2r}{T^2}$

Rearranging: $T^2 = \dfrac{4\pi^2r^3}{GM}$, i.e., $T^2 \propto r^3$ — confirming Kepler's third law.

$F = \dfrac{Gm_1m_2}{r^2} = \dfrac{6.67\times10^{-11}\times4\times9}{(0.5)^2}$

$F = \dfrac{6.67\times10^{-11}\times36}{0.25} = \dfrac{240.12\times10^{-11}}{0.25}$

$F \approx 960.5\times10^{-11} = \mathbf{9.6\times10^{-9}\text{ N}}$

Weight is inversely proportional to the square of the distance from the centre: $W \propto \dfrac{1}{r^2}$

$\dfrac{W_h}{W_{surface}} = \left(\dfrac{R}{2R}\right)^2 = \dfrac{1}{4}$

$W_h = \dfrac{72}{4} = \mathbf{18\text{ N}}$

$g_h = g\left(\dfrac{R}{R+h}\right)^2$. Given $g_h = 0.04g$:

$0.04 = \left(\dfrac{R}{R+h}\right)^2$

Taking square root: $0.2 = \dfrac{R}{R+h}$

$R + h = \dfrac{R}{0.2} = 5R \Rightarrow h = 5R - R = \mathbf{4R}$

At a height of 4 times Earth's radius, $g$ reduces to 4% of its surface value.

Orbital radius: $r = R + h = 6400 + 3600 = 10000\text{ km} = 1\times10^7\text{ m}$

Using $v_o = \sqrt{\dfrac{gR^2}{r}}$ (equivalent form using surface $g$):

$v_o = \sqrt{\dfrac{9.8\times(6.4\times10^6)^2}{1\times10^7}} = \sqrt{\dfrac{9.8\times4.096\times10^{13}}{1\times10^7}}$

$v_o = \sqrt{4.014\times10^7} \approx 6335\text{ m/s} \approx 6.34\text{ km/s}$

Time period: $T = \dfrac{2\pi r}{v_o} = \dfrac{2\pi(1\times10^7)}{6335} \approx \dfrac{6.28\times10^7}{6335} \approx 9917\text{ s} \approx 2.75\text{ hours}$

$v_e = \sqrt{\dfrac{2GM}{R}}$. Let planet's mass $M' = 2M$ and radius $R' = 2R$:

$v_e' = \sqrt{\dfrac{2G(2M)}{2R}} = \sqrt{\dfrac{2GM}{R}} = v_e$

The escape velocity remains unchanged at 11.2 km/s — since both mass and radius doubled proportionally, their ratio $M/R$ (which determines $v_e$) stays the same.

$g_h = g\left(\dfrac{R}{R+h}\right)^2 \Rightarrow \dfrac{4}{10} = \left(\dfrac{6400}{6400+h}\right)^2$

$0.4 = \left(\dfrac{6400}{6400+h}\right)^2 \Rightarrow \sqrt{0.4} \approx 0.632 = \dfrac{6400}{6400+h}$

$6400+h = \dfrac{6400}{0.632} \approx 10127\text{ km}$

$h \approx 10127 - 6400 = \mathbf{3727\text{ km}}$

By Kepler's third law: $T^2 \propto r^3$, so $\dfrac{T_A^2}{T_B^2} = \dfrac{r_A^3}{r_B^3} = \dfrac{r^3}{(4r)^3} = \dfrac{1}{64}$

$\dfrac{T_A}{T_B} = \sqrt{\dfrac{1}{64}} = \dfrac{1}{8}$

The ratio $T_A:T_B = 1:8$ — satellite B, orbiting 4 times farther out, takes 8 times as long to complete one orbit.

Orbital radius: $r = R + h = 6400 + 1000 = 7400\text{ km} = 7.4\times10^6\text{ m}$

$U = -\dfrac{GMm}{r} = -\dfrac{6.67\times10^{-11}\times6\times10^{24}\times500}{7.4\times10^6}$

Numerator: $6.67\times6\times500 = 20010$, so numerator $= 20010\times10^{13} = 2.001\times10^{17}$

$U = -\dfrac{2.001\times10^{17}}{7.4\times10^6} \approx \mathbf{-2.70\times10^{10}\text{ J}}$

At depth $d=R/2$: $g_d = g\left(1-\dfrac{d}{R}\right) = g\left(1-\dfrac{1}{2}\right) = \dfrac{g}{2}$

At height $h=R/2$: $g_h = g\left(\dfrac{R}{R+h}\right)^2 = g\left(\dfrac{R}{R+R/2}\right)^2 = g\left(\dfrac{R}{3R/2}\right)^2 = g\left(\dfrac{2}{3}\right)^2 = \dfrac{4g}{9}$

Ratio: $\dfrac{g_d}{g_h} = \dfrac{g/2}{4g/9} = \dfrac{1}{2}\times\dfrac{9}{4} = \mathbf{\dfrac{9}{8}}$

Using energy conservation between the surface and maximum height $h$ (where velocity = 0):

$\dfrac{1}{2}mv^2 - \dfrac{GMm}{R} = 0 - \dfrac{GMm}{R+h}$

Using $GM = gR^2$: $\dfrac{1}{2}v^2 = gR^2\left(\dfrac{1}{R}-\dfrac{1}{R+h}\right) = gR - \dfrac{gR^2}{R+h}$

Rearranging to solve for $h$: $\dfrac{gR^2}{R+h} = gR - \dfrac{v^2}{2}$

Substituting $v = 5000\text{ m/s}$, $g=9.8$, $R=6.4\times10^6\text{ m}$: $gR = 9.8\times6.4\times10^6 = 6.272\times10^7$; $\dfrac{v^2}{2} = \dfrac{25\times10^6}{2}=1.25\times10^7$

$gR - \dfrac{v^2}{2} = 6.272\times10^7 - 1.25\times10^7 = 5.022\times10^7$

$R+h = \dfrac{gR^2}{5.022\times10^7} = \dfrac{9.8\times(6.4\times10^6)^2}{5.022\times10^7} = \dfrac{4.014\times10^{14}}{5.022\times10^7} \approx 7.994\times10^6\text{ m}$

$h \approx 7.994\times10^6 - 6.4\times10^6 \approx \mathbf{1.59\times10^6\text{ m} \approx 1594\text{ km}}$