Gravitation — Easy Practice (Set 1)

Quick Recap — Gravitation Basics

• Universal law: $F = \dfrac{Gm_1 m_2}{r^2}$, with $G = 6.67\times10^{-11}$ N m$^2$/kg$^2$ — the force is always attractive.
• Acceleration due to gravity: $g = \dfrac{GM}{R^2} \approx 9.8$ m/s$^2$ at Earth's surface.
• Escape velocity: $v_e = \sqrt{2gR} = \sqrt{\dfrac{2GM}{R}} \approx 11.2$ km/s for Earth.
• Kepler's laws: orbits are ellipses; equal areas are swept in equal times (angular momentum); and $T^2 \propto r^3$.
• Gravity is a conservative force; $g$ is greatest at the poles and zero at the Earth's centre.

Beyond-NCERT JEE Formulae

Consolidated, exam-ready results that go past the plain NCERT statements — each tagged with when to use it and the trap examiners like to set.

1. Variation of $g$ (three regimes)

• Small height ($h\ll R$): $g_h=g\left(1-\dfrac{2h}{R}\right)$ — a linear approximation; use for mountain- or aircraft-scale heights.
• Any height: $g_h=g\left(\dfrac{R}{R+h}\right)^2$ — the exact inverse-square in distance from the centre; use when $h$ is comparable with $R$.
• Depth: $g_d=g\left(1-\dfrac{d}{R}\right)$ — falls linearly to zero at the centre of a uniform Earth.
• Latitude: $g_\lambda=g-\omega^2R\cos^2\lambda$ — the spin cuts $g$ most at the equator ($\lambda=0$) and not at all at the poles.

[JEE Tip] Height weakens $g$ TWICE as fast as depth: compare the factor $\dfrac{2h}{R}$ against $\dfrac{d}{R}$. So the depth matching a small height $h$ is $d=2h$.

2. Gravitational field $E$ and potential $V$

Outside any spherical body, or for a point mass ($r\ge R$): $$E=\dfrac{GM}{r^2},\qquad V=-\dfrac{GM}{r}$$

• Solid sphere, inside ($r<R$): field $E=\dfrac{GMr}{R^3}$ grows linearly, while $V=-\dfrac{GM\left(3R^2-r^2\right)}{2R^3}$; at the centre $V_c=-\dfrac{3GM}{2R}$.
• Thin shell, inside ($r<R$): field $E=0$, yet the potential stays flat at $V=-\dfrac{GM}{R}$ (its surface value).
• Ring, on the axis at distance $x$: field $E=\dfrac{GMx}{\left(R^2+x^2\right)^{3/2}}$ points along the axis, with $V=-\dfrac{GM}{\sqrt{R^2+x^2}}$.

[JEE Tip] Field is a VECTOR (resolve and add components); potential is a SCALAR (add signed values, ignore direction). Inside a shell $E=0$ but $V\ne0$ — a null field never forces a null potential. The two are linked by $E=-\dfrac{dV}{dr}$.

3. Escape and orbital speed

• Escape speed: $v_e=\sqrt{2gR}=\sqrt{\dfrac{2GM}{R}}$ — independent of the body's mass and of the launch angle, since energy is a scalar.
• Orbital speed at radius $r$: $v_o=\sqrt{\dfrac{GM}{r}}$; skimming the surface this is $v_o=\sqrt{gR}$.
• Master link: $v_e=\sqrt2\,v_o$ — escaping needs only about 41 percent more speed than circling.

[JEE Tip] When the mass is missing but the density $\rho$ is given, switch to $v_e=R\sqrt{\dfrac{8\pi G\rho}{3}}$ and $g=\dfrac{4}{3}\pi G\rho R$ — both follow from $M=\dfrac{4}{3}\pi R^3\rho$.

4. Energy of a satellite (circular orbit, radius $r$)

$$KE=\dfrac{GMm}{2r},\qquad PE=-\dfrac{GMm}{r},\qquad E=-\dfrac{GMm}{2r}$$

• The signs lock into a fixed pattern: $KE=-E$ together with $PE=2E=-2\,KE$.
• Binding energy — the least energy to send it from the orbit to rest at infinity — equals $\dfrac{GMm}{2r}$; every BOUND orbit has $E<0$.
• To break free FROM the orbit, raise the speed by the factor $\sqrt2$, that is $\Delta v=\left(\sqrt2-1\right)v_o$.

[JEE Tip] A satellite dragged by the thin upper atmosphere actually SPEEDS UP: as $r$ shrinks, $KE=\dfrac{GMm}{2r}$ rises. The energy lost to friction is drawn out of the deeper potential well.

5. Kepler's third law and the geostationary orbit

• $T^2=\dfrac{4\pi^2}{GM}\,r^3$, so $T^2\propto r^3$ for one central body; inverted, $r\propto T^{2/3}$.
• A grazing circular orbit depends on density alone: $T=\sqrt{\dfrac{3\pi}{G\rho}}$ (about 84 min at Earth's density, whatever the planet's size).
• Geostationary satellite: a period of 24 h fixes $r\approx4.2\times10^4$ km (roughly 36000 km of height); it must ride over the EQUATOR moving west to east.

[JEE Tip] For a two-body BINARY the law is set by the SUM of the masses: $T=2\pi\sqrt{\dfrac{d^3}{G\left(m_1+m_2\right)}}$ — never by one star alone.

Solved Examples — Beyond-NCERT Formulae

Example 1 — Same offset, up versus down. A planet has surface gravity $g$. Find the acceleration due to gravity at a height $h=\dfrac{R}{2}$ and at a depth $d=\dfrac{R}{2}$, and say which is larger.

• Height (exact law): $g_h=g\left(\dfrac{R}{R+h}\right)^2=g\left(\dfrac{R}{3R/2}\right)^2=g\left(\dfrac{2}{3}\right)^2=\dfrac{4g}{9}\approx0.444g$.
• Depth (linear law): $g_d=g\left(1-\dfrac{d}{R}\right)=g\left(1-\dfrac{1}{2}\right)=\dfrac{g}{2}=0.5g$.

Answer: $g_h=\dfrac{4g}{9}\approx0.44g$ while $g_d=0.5g$, so gravity survives BETTER going down. Beware — the small-height rule $g\left(1-\dfrac{2h}{R}\right)$ does not apply here, because $h$ is not small compared with $R$.

Example 2 — Orbital versus escape speed. A moon has $g=8$ m/s$^2$ at its surface and radius $R=4000$ km. Find the speed of a surface-skimming satellite and the escape speed.

• Orbital: $v_o=\sqrt{gR}=\sqrt{(8)\left(4\times10^{6}\right)}=\sqrt{3.2\times10^{7}}=5.66\times10^{3}$ m/s.
• Escape: $v_e=\sqrt{2}\,v_o=\sqrt{2gR}=\sqrt{6.4\times10^{7}}=8.0\times10^{3}$ m/s.

Answer: $v_o\approx5.66$ km/s and $v_e\approx8.0$ km/s, standing in the ratio $1:\sqrt2$. Writing $\sqrt{2gR}$ for the orbital speed is the classic slip — that expression is the escape speed, larger by $\sqrt2$.

Example 3 — Total energy and period of a satellite. A satellite of mass $m=200$ kg orbits the Earth at radius $r=3R$, with $g=9.8$ m/s$^2$ and $R=6400$ km. Find its total mechanical energy and its orbital period.

• Work with $GM=gR^2=(9.8)\left(6.4\times10^{6}\right)^2=4.01\times10^{14}$ in SI units, and $r=3R=1.92\times10^{7}$ m.
• Energy: $E=-\dfrac{GMm}{2r}=-\dfrac{\left(4.01\times10^{14}\right)(200)}{2\left(1.92\times10^{7}\right)}=-2.09\times10^{9}$ J.
• Period: $T=2\pi\sqrt{\dfrac{r^3}{GM}}=2\pi\sqrt{\dfrac{\left(1.92\times10^{7}\right)^3}{4.01\times10^{14}}}=2\pi(4199)=2.64\times10^{4}$ s.

Answer: $E\approx-2.09\times10^{9}$ J, negative as any bound orbit must be, and $T\approx2.64\times10^{4}$ s, about 7.3 h. Quoting the potential energy $-\dfrac{GMm}{r}$ alone would double the magnitude of the energy.

Example 4 — Kepler ratio. Two satellites circle the same planet with orbital radii in the ratio $1:9$. Find the ratio of their periods.

• Kepler's third law gives $T\propto r^{3/2}$, so $\dfrac{T_2}{T_1}=\left(\dfrac{r_2}{r_1}\right)^{3/2}$.
• Substituting the radius ratio: $\dfrac{T_2}{T_1}=9^{3/2}=\left(9^{1/2}\right)^3=3^3=27$.

Answer: the periods are in the ratio $1:27$. Taking $T\propto r$ would wrongly give $1:9$; the exponent is $\dfrac{3}{2}$, not $1$.

Example 5 — Field and potential of a solid sphere. For a uniform solid sphere of mass $M$ and radius $R$, find the field and the potential at an inside point $r=\dfrac{R}{2}$ and at an outside point $r=2R$.

• Inside ($r=\tfrac R2$): field $E=\dfrac{GMr}{R^3}=\dfrac{GM\left(R/2\right)}{R^3}=\dfrac{GM}{2R^2}$, and $V=-\dfrac{GM\left(3R^2-r^2\right)}{2R^3}=-\dfrac{GM\left(3R^2-R^2/4\right)}{2R^3}=-\dfrac{11GM}{8R}$.
• Outside ($r=2R$): field $E=\dfrac{GM}{r^2}=\dfrac{GM}{4R^2}$, and $V=-\dfrac{GM}{r}=-\dfrac{GM}{2R}$.

Answer: inside, $E=\dfrac{GM}{2R^2}$ with $V=-\dfrac{11GM}{8R}$; outside, $E=\dfrac{GM}{4R^2}$ with $V=-\dfrac{GM}{2R}$. Note the potential is deepest at the CENTRE, where it is $-\dfrac{3GM}{2R}$, not at the surface.

Example 6 — Radius of a geostationary orbit. A satellite is to hover above one point on the equator, so its period is $T=24$ h. Taking $g=9.8$ m/s$^2$ and $R=6400$ km, find its orbital radius.

• Rearrange Kepler's law for the radius: $r^3=\dfrac{GM\,T^2}{4\pi^2}$, with $GM=gR^2=4.01\times10^{14}$ and $T=86400$ s.
• Compute: $r^3=\dfrac{\left(4.01\times10^{14}\right)\left(86400\right)^2}{4\pi^2}=7.59\times10^{22}$, so $r=\left(7.59\times10^{22}\right)^{1/3}=4.23\times10^{7}$ m.

Answer: $r\approx4.23\times10^{4}$ km from the Earth's centre, a height of about 36000 km. The famous 36000 km figure is the ALTITUDE; the orbital radius, measured from the centre, is one Earth radius more.

Example 7 — Binding energy and boosting an orbit. A satellite of mass $m=100$ kg orbits the Earth at radius $2R$, with $g=9.8$ m/s$^2$ and $R=6400$ km. Find (a) its binding energy and (b) the extra energy needed to lift it to a circular orbit of radius $3R$.

• Binding energy at $2R$: $\dfrac{GMm}{2(2R)}=\dfrac{gR^2m}{4R}=\dfrac{mgR}{4}=\dfrac{(100)(9.8)\left(6.4\times10^{6}\right)}{4}=1.57\times10^{9}$ J.
• Raising $2R\to3R$: $\Delta E=\left(-\dfrac{GMm}{6R}\right)-\left(-\dfrac{GMm}{4R}\right)=\dfrac{GMm}{12R}=\dfrac{mgR}{12}=5.23\times10^{8}$ J.

Answer: the binding energy is $\approx1.57\times10^{9}$ J and the transfer needs $\approx5.23\times10^{8}$ J. Climbing to the higher orbit costs energy even though the orbital SPEED there is smaller — the potential well pays the kinetic loss back twice over.

Example 8 — Effective gravity at a latitude. The Earth ($R=6400$ km, day 24 h) turns once about its axis. By how much does this spin reduce the effective $g$ at latitude $\lambda=45^\circ$?

• Angular speed: $\omega=\dfrac{2\pi}{86400}=7.27\times10^{-5}$ rad/s, giving the equatorial reduction $\omega^2R=\left(7.27\times10^{-5}\right)^2\left(6.4\times10^{6}\right)=3.38\times10^{-2}$ m/s$^2$.
• At latitude $45^\circ$: reduction $=\omega^2R\cos^2\lambda=\left(3.38\times10^{-2}\right)\left(\cos^245^\circ\right)=\left(3.38\times10^{-2}\right)(0.5)=1.69\times10^{-2}$ m/s$^2$.

Answer: the effective $g$ drops by about $1.7\times10^{-2}$ m/s$^2$ at $45^\circ$ — exactly half the equatorial reduction, since $\cos^245^\circ=\tfrac12$ — leaving $g_\lambda\approx9.78$ m/s$^2$. The cosine enters SQUARED; using it only once overstates the effect.