Gravitation

Newton’s law \(F = Gm_1m_2/r^2\) has five features that NEET tests directly.

1. Always attractive and central. The force acts along the line joining the two centres.
2. Equal and opposite. The Earth pulls an apple with the same force the apple pulls the Earth. The accelerations differ because \(a = F/m\): the apple accelerates a lot, the Earth almost not at all.
3. Independent of the medium. Placing a third body or a wall between two masses does not shield or change the force between them. Each pair interacts on its own.
4. Universal constant. \(G = 6.67\times10^{-11}\ \text{N m}^2\,\text{kg}^{-2}\), with dimensions \([M^{-1}L^3T^{-2}]\). \(G\) is the same everywhere. The acceleration \(g\) is not; it depends on where you are.
5. Very weak. Two 1 kg masses 1 m apart attract with only about \(6.7\times10^{-11}\) N. Gravity dominates in astronomy only because planets are huge and electrically neutral.

Fields and forces from several masses add as vectors. A few set-ups appear again and again.

• Null point between two masses. Between masses \(m_1\) and \(m_2\) a distance \(d\) apart, the fields cancel where \(Gm_1/x^2 = Gm_2/(d-x)^2\). This gives \(x = d\sqrt{m_1}/(\sqrt{m_1}+\sqrt{m_2})\) from \(m_1\). The null point sits closer to the smaller mass.
• Symmetric arrangements. Equal masses at the corners of an equilateral triangle, square or regular polygon give zero field at the centre. The field at the centre of a uniform ring is also zero.
• Remove one mass. If one mass is taken away from a symmetric set, the field at the centre equals the field of the missing mass, but pointing away from where it was. The remaining masses no longer have a partner to cancel them.
• Force on one corner mass. For three equal masses at the corners of an equilateral triangle of side \(a\), each pair force is \(Gm^2/a^2\) at 60° to each other, so the net force on one mass is \(\sqrt3\,Gm^2/a^2\).

At the surface, the weight \(mg\) is the gravitational force \(GMm/R^2\). So \(g = GM/R^2\). Writing \(M = \tfrac43\pi R^3\rho\) gives \(g = \tfrac43\pi G\rho R\).

Use this to compare planets. If two planets have the same density, \(g \propto R\). If they have the same mass, \(g \propto 1/R^2\). Pick the form that holds the given quantity fixed, then take the ratio.

At height \(h\) the body is \(R+h\) from the centre, and all of Earth’s mass lies inside that radius.

1. \(g_h = \dfrac{GM}{(R+h)^2}\) and \(g = \dfrac{GM}{R^2}\), so \(g_h = g\left(\dfrac{R}{R+h}\right)^2 = g\left(1+\dfrac hR\right)^{-2}\).
2. For \(h \ll R\), the binomial approximation \((1+x)^{-2}\approx 1-2x\) gives \(g_h \approx g\left(1-\dfrac{2h}{R}\right)\).
3. The fractional decrease is \(\Delta g/g \approx 2h/R\). At 64 km above Earth (\(R = 6400\) km) this is 2%.

Use the exact form when \(h\) is comparable to \(R\). At \(h = R\) the exact value is \(g/4\); the approximation would give \(-g\), which is nonsense.

Picture a body in a mine at depth \(d\). Model Earth as a uniform sphere.

1. The shell of rock above the body (outer thickness \(d\)) gives zero net field inside it (shell theorem).
2. Only the inner sphere of radius \(R-d\) pulls. Its mass is \(M' = M\dfrac{(R-d)^3}{R^3}\), because mass scales with volume at fixed density.
3. \(g_d = \dfrac{GM'}{(R-d)^2} = \dfrac{GM(R-d)}{R^3} = g\left(1-\dfrac dR\right)\). This result is exact for a uniform sphere, not an approximation.
4. At the centre (\(d = R\)), \(g = 0\).

For the same small distance, going up reduces \(g\) twice as fast as going down: \(2h/R\) against \(d/R\). So \(g\) at a small height \(h\) equals \(g\) at depth \(d = 2h\).

Plot \(g\) against distance \(r\) from the centre. Inside, \(g = gr/R\) rises as a straight line from zero. Outside, \(g = gR^2/r^2\) falls as an inverse square. The peak is at the surface. Moving from the surface in either direction lowers \(g\).

The mass of a body never changes with location. Only its weight \(mg\) changes.

1. A point P at latitude \(\lambda\) is a distance \(r = R\cos\lambda\) from the spin axis.
2. To move in that circle, it needs a centripetal acceleration \(\omega^2R\cos\lambda\), directed toward the axis (horizontally inward in the figure).
3. Resolve this along the line to the Earth’s centre: the component is \(\omega^2R\cos\lambda\times\cos\lambda = \omega^2R\cos^2\lambda\).
4. Gravity \(g\) must supply this. What is left to press on the scale is \(g' \approx g - \omega^2R\cos^2\lambda\). A small sideways part (and an \(\omega^4\) term) is neglected.

Checks: at the equator \(g' = g - \omega^2R\), at the poles \(g' = g\). With \(\omega^2R\approx0.034\) m/s², the effect is small but measurable.

The Earth bulges at the equator: the equatorial radius is about 21 km more than the polar radius. A point at the equator is therefore farther from the centre, which lowers \(GM/R^2\) as well. Both effects work the same way: measured \(g\) is about 9.78 m/s² at the equator and about 9.83 m/s² at the poles.

So a body weighs slightly more at the poles. Its mass is the same everywhere.

A body at the equator would feel weightless if \(\omega^2R = g\), that is, \(\omega = \sqrt{g/R}\). With \(g = 10\) m/s² and \(R = 6.4\times10^6\) m, \(\omega = 1.25\times10^{-3}\) rad/s, giving a day of \(2\pi/\omega\approx5030\) s, about 1.4 h, roughly 17 times faster than now. This is the same as the period of a satellite skimming the surface: a body at the equator would then be “in orbit” while standing on the ground.

For other latitudes the approximate formula is not reliable at such large \(\omega\), so NEET uses only the equator for this question.

Potential energy at distance \(r\) is the work done by an external agent to bring \(m\) slowly from infinity to \(r\).

1. Measure \(x\) outward from \(M\). Gravity on \(m\) is \(-GMm/x^2\) (inward), so the agent applies \(+GMm/x^2\) (outward) to move it without speeding up.
2. \(W = \displaystyle\int_\infty^r \frac{GMm}{x^2}\,dx = \left[-\frac{GMm}{x}\right]_\infty^r = -\frac{GMm}{r}\). So \(U(r) = -\dfrac{GMm}{r}\).
3. A quicker check: gravity does positive work as \(m\) falls inward, so the system loses potential energy. Starting from zero at infinity, \(U\) must be negative everywhere else.

The potential is \(V = U/m = -GM/r\) (J/kg). It is a scalar, so for several masses you simply add \(-Gm_i/r_i\) with signs; no directions are needed.

Lift a mass \(m\) from the surface to height \(h\):

\(\Delta U = GMm\left(\dfrac1R-\dfrac1{R+h}\right) = \dfrac{GMm\,h}{R(R+h)} = \dfrac{mgh}{1+h/R}\).

For \(h\ll R\) this reduces to \(mgh\). For \(h = R\) it gives \(mgR/2\), not \(mgR\). The familiar \(mgh\) is a near-surface approximation that assumes \(g\) stays constant.

For several particles, the potential energy of the system is the sum over every pair: \(U = -\sum Gm_im_j/r_{ij}\). Three equal masses on an equilateral triangle of side \(a\) have three pairs, so \(U = -3Gm^2/a\). The work needed to pull them apart to infinity is \(+3Gm^2/a\).

Field is the negative slope of potential: \(E = -dV/dr\). Where \(V\) is flat (inside a shell, or at a maximum or minimum of \(V\)), the field is zero, even though \(V\) itself is not zero. On the axis of a ring of mass \(M\) and radius \(a\), \(V = -GM/\sqrt{a^2+x^2}\); at the centre \(V = -GM/a\) while the field is zero.

For a circular orbit of radius \(r = R + h\), gravity provides the centripetal force:

1. \(\dfrac{mv^2}{r} = \dfrac{GMm}{r^2}\), so \(v_o = \sqrt{\dfrac{GM}{r}} = \sqrt{\dfrac{gR^2}{R+h}}\).
2. Period: \(T = \dfrac{2\pi r}{v_o} = 2\pi\sqrt{\dfrac{r^3}{GM}}\).
3. Close to the surface (\(h \ll R\)): \(v_o \approx \sqrt{gR}\approx 8\) km/s and \(T \approx 2\pi\sqrt{R/g}\approx 84\) minutes with \(g = 10\) m/s² and \(R = 6400\) km.

The satellite mass \(m\) cancels. Two satellites in the same orbit have the same speed and period whatever their masses.

With \(v_o^2 = GM/r\):

• Kinetic energy \(K = \tfrac12mv_o^2 = +\dfrac{GMm}{2r}\).
• Potential energy \(U = -\dfrac{GMm}{r}\).
• Total energy \(E = K + U = -\dfrac{GMm}{2r}\).

So \(K : U : E = 1 : -2 : -1\). The binding energy (energy needed to free the satellite from its orbit) is \(+GMm/2r = K\).

A higher orbit has less kinetic energy but more total energy (\(E\) is less negative). That is why moving a satellite outward needs energy: \(\Delta E = \dfrac{GMm}{2}\left(\dfrac1{r_1}-\dfrac1{r_2}\right)\). Launching from the surface (ignoring Earth’s rotation) into an orbit of radius \(r\) needs \(E_{\rm orbit} - U_{\rm surface} = GMm\left(\dfrac1R - \dfrac1{2r}\right)\).

Scaling rules for circular orbits around the same planet: \(v_o \propto r^{-1/2}\), \(T \propto r^{3/2}\), \(K \propto 1/r\), \(|E| \propto 1/r\), angular momentum \(L = m\sqrt{GMr} \propto r^{1/2}\).

• If gravity were switched off, the satellite would fly off along the tangent with speed \(v_o\).
• If its speed is raised to \(\sqrt2\,v_o\) (about 41.4% more), it escapes.
• Speed between \(v_o\) and \(\sqrt2\,v_o\): an ellipse, with the starting point as the nearest point. Speed less than \(v_o\): an ellipse with the starting point as the farthest point; near the Earth it may hit the ground.

A scale measures the normal force \(N\) between you and it. For a person in a lift with upward acceleration \(a\) (taking up as positive):

• At rest or moving at constant velocity: \(N = mg\).
• Accelerating up (or slowing while moving down): \(N = m(g+a)\), you feel heavier.
• Accelerating down (or slowing while moving up): \(N = m(g-a)\), you feel lighter.
• Cable snaps, free fall: \(a = g\) down, \(N = 0\), weightless.
• Accelerating down faster than \(g\): you would be pressed against the ceiling.

Inside an orbiting satellite, every object falls with the same acceleration as the satellite. So:

• A spring balance hung from the ceiling reads zero for any object.
• A simple pendulum does not oscillate; with no effective gravity it stays wherever it is put.
• Water does not pour out of a tilted glass, and a candle flame becomes nearly spherical because hot gases no longer rise.
• A body released from the satellite does not fall to the Earth. It keeps the satellite’s velocity, so it stays in the same orbit alongside.
• Mass can still be measured, by an inertial method: put the body on a spring and time its oscillations, then use \(m = kT^2/4\pi^2\).

A rotating space station can press astronauts against its outer wall. The wall provides the centripetal force: \(N = m\omega^2r\). Choosing \(\omega^2r = g\) makes the reading the same as on Earth. For a station of radius 100 m, \(\omega = \sqrt{10/100}\approx0.32\) rad/s, about one turn every 20 s.

A body launched with speed \(v\) from a planet’s surface just escapes if it reaches infinity with zero speed. Total energy at infinity is then zero, so at launch:

1. \(\tfrac12mv_e^2 - \dfrac{GMm}{R} = 0\).
2. \(v_e = \sqrt{\dfrac{2GM}{R}} = \sqrt{2gR}\).
3. For Earth with \(g = 9.8\) m/s² and \(R = 6.4\times10^6\) m, \(v_e = 11.2\) km/s. With \(g = 10\) it is about 11.3 km/s.
4. Using \(M = \tfrac43\pi R^3\rho\): \(v_e = R\sqrt{\dfrac{8\pi G\rho}{3}}\), so at equal density \(v_e\propto R\).

The launch direction does not matter (ignoring air and rotation), because energy is a scalar. The mass of the body cancels.

• Slower (vertical launch). It rises to height \(h\) where its speed is zero: \(\tfrac12mv^2 - \dfrac{GMm}{R} = -\dfrac{GMm}{R+h}\). This gives \(h = \dfrac{Rv^2}{v_e^2 - v^2}\). At \(v = v_e/2\), \(h = R/3\). At \(v = v_e/\sqrt2\), \(h = R\).
• Faster. It escapes with leftover speed \(v_\infty = \sqrt{v^2 - v_e^2}\) far away. Launched at \(2v_e\), it keeps \(\sqrt3\,v_e\).
• From an orbit. At radius \(r\), \(v_e = \sqrt{2GM/r} = \sqrt2\,v_o\). A satellite needs only \(\sqrt2 - 1\approx 41.4\%\) more speed to escape.

Escape speed on the Moon is only about 2.4 km/s. Gas molecules at the Moon’s daytime temperatures move with average speeds that are a sizeable fraction of this, and the fastest molecules in the distribution exceed it. Over long times, the gas leaks away. Earth’s larger escape speed keeps nitrogen and oxygen, though even Earth has lost most of its light hydrogen and helium.

1. Law of orbits. Each planet moves in an ellipse with the Sun at one focus (not at the centre).
2. Law of areas. The line from the Sun to the planet sweeps equal areas in equal times. The planet moves fastest at perihelion (nearest point) and slowest at aphelion (farthest point).
3. Law of periods. \(T^2\propto a^3\), where \(a\) is the semi-major axis, \(a = (r_{\min}+r_{\max})/2\).

Third law for a circular orbit. Gravity provides the centripetal force: \(\dfrac{GMm}{r^2} = m\left(\dfrac{2\pi}{T}\right)^2r\). Rearranging, \(T^2 = \dfrac{4\pi^2}{GM}r^3\). For an ellipse, the same result holds with \(r\) replaced by \(a\).

Second law from angular momentum. In a short time \(dt\), the radius sweeps a thin triangle of area \(dA = \tfrac12 r\,(v\,dt)\sin\theta\). So \(\dfrac{dA}{dt} = \dfrac{|\vec r\times\vec v|}{2} = \dfrac{L}{2m}\). Gravity is a central force, so its torque about the Sun is zero and \(L\) is constant. Hence the areal speed is constant. At perihelion and aphelion the velocity is perpendicular to the radius, giving \(v_pr_p = v_ar_a\).

A geostationary satellite appears fixed in the sky. It needs all of these: a circular orbit in the equatorial plane, a period of 24 h (strictly one sidereal day, about 23 h 56 min), and motion from west to east, like the Earth’s spin. From \(T^2\propto r^3\) its orbit radius is about 42,000 km (about 6.6 Earth radii), a height of about 36,000 km, with speed about 3.1 km/s. It is used for communication and weather monitoring.

A polar satellite passes over the poles at a low height (roughly 500–800 km) with a period of about 100 minutes. The Earth turns beneath it, so it scans the whole surface. It is used for remote sensing and mapping.

For a uniform thin shell of mass \(M\) and radius \(R\):

• Outside (\(r > R\)): it acts like a point mass at the centre. \(E = GM/r^2\), \(V = -GM/r\).
• Inside (\(r < R\)): \(E = 0\) everywhere. The potential is constant and equal to its surface value, \(V = -GM/R\).

The field jumps from 0 just inside to \(GM/R^2\) just outside. The potential is continuous, with a flat piece inside joined to the \(-GM/r\) curve outside. A body inside a shell feels no force from the shell, but work is still needed to take it to infinity.

• Outside: same as a point mass, \(E = GM/r^2\), \(V = -GM/r\).
• Inside: only the mass within radius \(r\) pulls, \(M r^3/R^3\). So \(E = GMr/R^3\), rising linearly from zero at the centre to \(GM/R^2\) at the surface.
• Inside potential: \(V = -\dfrac{GM}{2R^3}(3R^2 - r^2)\). It is a downward parabola, most negative at the centre, \(V_{\rm centre} = -\dfrac{3GM}{2R} = 1.5\,V_{\rm surface}\).

Unlike the shell, the solid sphere’s field graph is continuous with a peak at the surface. The field equals half its surface value at two places: \(r = R/2\) inside and \(r = \sqrt2R\) outside.

Moving a mass \(m\) from the centre of a uniform sphere to its surface raises its energy by \(m(V_{\rm surface} - V_{\rm centre}) = m\left(-\dfrac{GM}{R}+\dfrac{3GM}{2R}\right) = \dfrac{GMm}{2R}\). Reverse it: a body dropped from the surface down a frictionless tunnel through the centre reaches the centre with \(\tfrac12mv^2 = GMm/2R\), so \(v = \sqrt{GM/R} = \sqrt{gR}\), about 8 km/s for Earth. The Oscillations chapter shows this motion is simple harmonic.

Earth’s rotation and latitude also change the measured \(g\). That is treated in its own section, “Earth’s rotation and latitude”.