Gravitation — Universal Law, Orbital Mechanics, Satellites

The Force That Holds Everything Together

Gravitation is the attractive force between any two masses in the universe. Newton’s universal law of gravitation explains why apples fall, why the Moon orbits Earth, and why planets orbit the Sun. For Class 9, we cover the basics. For Class 11, we add Kepler’s laws, orbital mechanics, and escape velocity.

CBSE Class 9 gives 3-4 marks. Class 11 gives 5-7 marks. JEE Main tests 1-2 questions, often on satellite motion or gravitational PE. The chapter is formula-heavy but the formulas are elegant — most follow from one equation ( $F = GMm/r^2$ ) and the energy conservation principle.

Core Concepts

Kepler’s three laws of planetary motion

Before Newton, Johannes Kepler (1609-1619) described planetary orbits empirically:

Law of Orbits: Every planet moves in an ellipse with the Sun at one focus.
Law of Areas: A line from the Sun to a planet sweeps equal areas in equal times — the planet moves faster when closer to the Sun (perihelion) and slower when farther (aphelion).
Law of Periods: $T^2 \propto a^3$ — the square of the orbital period is proportional to the cube of the semi-major axis.

Newton later proved all three laws are consequences of his universal law of gravitation and the laws of motion.

Newton’s law of universal gravitation

F = \frac{Gm_1 m_2}{r^2}

$G = 6.674 \times 10^{-11}$ N m $^2$ /kg $^2$ (universal gravitational constant). Always attractive. Acts along the line joining the centres of mass. Obeys the superposition principle.

Key properties: (1) Universal — every mass attracts every other mass. (2) Always attractive — there is no gravitational repulsion. (3) Independent of the medium — unlike electrostatic force, gravity cannot be shielded. (4) Inverse-square — doubling the distance reduces the force by 4 times.

Cavendish experiment (1798): First measurement of $G$ using a torsion balance with lead spheres. This allowed calculation of Earth’s mass: $M_E = gR^2/G$ .

Acceleration due to gravity

At the surface: $g = \frac{GM}{R^2} \approx 9.8$ m/s $^2$

At height $h$ above surface: $g_h = g\left(\frac{R}{R+h}\right)^2$

For $h \ll R$ : $g_h \approx g\left(1 - \frac{2h}{R}\right)$ (approximate formula)

At depth $d$ below surface: $g_d = g\left(1 - \frac{d}{R}\right)$

At the centre ( $d = R$ ): $g = 0$

Important: $g$ decreases both above and below the surface, but for different reasons:

Above: you are farther from the centre, so $r$ in $GM/r^2$ increases
Below: only the mass within your radius contributes (shell theorem), and that mass decreases with depth

$g$ also varies with latitude due to Earth’s rotation (centrifugal effect) and non-spherical shape (Earth is an oblate spheroid — wider at the equator).

Location	$g$ value	Why
Poles	Maximum (9.83 m/s $^2$ )	No centrifugal reduction, closer to centre
Equator	Minimum (9.78 m/s $^2$ )	Maximum centrifugal reduction, farther from centre

Gravitational potential energy

PE = -\frac{GMm}{r}

Negative because we define PE = 0 at infinity. Closer to the mass (smaller $r$ ) means more negative PE — the object is more tightly bound.

Near the surface (for small heights $h$ ): $PE \approx mgh$ (the familiar form from Class 9).

Escape velocity

The minimum velocity needed to escape a gravitational field completely (reach $r = \infty$ with zero velocity).

v_e = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}

For Earth: $v_e = \sqrt{2 \times 9.8 \times 6.4 \times 10^6} \approx 11.2$ km/s.

Note: $v_e = \sqrt{2} \times v_o$ (escape velocity is $\sqrt{2}$ times the orbital velocity at the surface).

Key facts: Escape velocity does not depend on the mass or direction of the object (only on $M$ and $R$ of the planet). On the Moon, $v_e \approx 2.4$ km/s — low enough that the Moon has no atmosphere (gas molecules can escape).

Orbital mechanics

Orbital velocity at radius $r$ : $v_o = \sqrt{\frac{GM}{r}}$

At surface ( $r \approx R$ ): $v_o = \sqrt{gR} \approx 7.9$ km/s for Earth.

Time period: $T = 2\pi\sqrt{\frac{r^3}{GM}} = \frac{2\pi r}{v_o}$

This is Kepler’s third law derived from Newton’s laws.

Energy of an orbiting satellite

$KE = \frac{GMm}{2r}$ (always positive)

$PE = -\frac{GMm}{r}$ (always negative)

Total energy $E = KE + PE = -\frac{GMm}{2r}$ (always negative for a bound orbit)

Note: $|PE| = 2 \times KE$ and $|E| = KE$ . Total energy is negative — the satellite is bound.

To escape, a satellite needs extra energy equal to its binding energy: $\Delta E = +\frac{GMm}{2r}$ .

Types of orbits and satellites

Type	Height	Period	Use
LEO (Low Earth Orbit)	200-2000 km	~90 min	ISS, Earth observation, imaging
MEO (Medium Earth Orbit)	2000-36000 km	2-24 hrs	GPS (~20,200 km, ~12 hr period)
GEO (Geostationary)	~36,000 km	24 hrs	Communication, weather (stays fixed above one point)
Polar orbit	~800 km	~100 min	Mapping (covers entire Earth as it rotates)

Geostationary conditions: orbit must be circular, equatorial, and at 36,000 km altitude with period = 24 hours. Only one such orbit exists.

Weightlessness

An astronaut in orbit is not beyond Earth’s gravity — at ISS altitude (~400 km), $g$ is still about 8.7 m/s $^2$ (only ~11% less than surface). The astronaut feels weightless because both the astronaut and the spacecraft are in free fall — falling toward Earth at the same rate. The normal force (which we perceive as weight) is zero.

True weightlessness only occurs at the centre of a gravitational field or infinitely far from all masses.

Solved Examples

Example 1 (Easy — CBSE Class 9)

Find the gravitational force between two 50 kg masses placed 1 m apart.

$F = G \frac{m_1 m_2}{r^2} = 6.674 \times 10^{-11} \times \frac{50 \times 50}{1^2} = 6.674 \times 10^{-11} \times 2500 = \mathbf{1.67 \times 10^{-7}}$ N.

This is an incredibly tiny force — gravity is the weakest of the four fundamental forces. We only feel it because Earth is so massive.

Example 2 (Medium — JEE Main)

Given $g = 10$ m/s $^2$ , $R = 6400$ km = $6.4 \times 10^6$ m.

$v_e = \sqrt{2gR} = \sqrt{2 \times 10 \times 6.4 \times 10^6} = \sqrt{1.28 \times 10^8} = 1.131 \times 10^4$ m/s $= \mathbf{11.3}$ km/s.

This is about 40 times the speed of sound. Rockets achieve this gradually, not instantaneously.

Example 3 (Hard — JEE Main)

A satellite orbits at height $h = R$ above Earth’s surface. Find its speed and time period in terms of $g$ and $R$ .

$r = R + h = R + R = 2R$

$v = \sqrt{\frac{GM}{2R}} = \sqrt{\frac{gR^2}{2R}} = \sqrt{\frac{gR}{2}}$

$T = 2\pi\sqrt{\frac{(2R)^3}{GM}} = 2\pi\sqrt{\frac{8R^3}{gR^2}} = 2\pi\sqrt{\frac{8R}{g}}$

Example 4 (Medium — CBSE)

$g_h = g\left(\frac{R}{R+h}\right)^2 = g\left(\frac{R}{R+R}\right)^2 = g\left(\frac{1}{2}\right)^2 = \frac{g}{4} = \mathbf{2.45}$ m/s $^2$ .

At height R, gravity is one-quarter of the surface value. Note: you must use the exact formula, not the approximation (which only works for $h \ll R$ ).

Example 5 (Hard — JEE)

For small $h$ : $g_h \approx g(1 - 2h/R)$ and $g_d = g(1 - d/R)$ .

Setting them equal: $1 - 2h/R = 1 - d/R$ , so $d = 2h$ .

At depth $d = 2h$ , $g$ is the same as at height $h$ . Going underground reduces $g$ linearly; going up reduces it quadratically (for small distances). So for the same reduction in $g$ , you need to go twice as deep as high.

Example 6 (Application)

A satellite of mass $m$ is in orbit at radius $r_1$ . To move it to $r_2 > r_1$ :

$\Delta E = E_2 - E_1 = -\frac{GMm}{2r_2} - \left(-\frac{GMm}{2r_1}\right) = \frac{GMm}{2}\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$

This is the minimum energy input needed (provided by rocket thrust). The satellite moves to a higher orbit but actually slows down ( $v \propto 1/\sqrt{r}$ ) — a counterintuitive result.

Common Mistakes to Avoid

Using $r = h$ instead of $r = R + h$ . The distance in gravitational formulas is from the centre of Earth, not from the surface. A satellite at height $h$ has $r = R + h$ . This is the single most common mistake in gravitation problems.

Forgetting that PE is negative. Gravitational PE $= -GMm/r$ . It is zero at infinity and becomes more negative closer to the mass. A common error is writing PE as positive, which flips all energy calculations.

Confusing orbital velocity with escape velocity. $v_e = \sqrt{2} \times v_o$ at the same radius. Escape velocity is about 41% more than orbital velocity. An orbiting satellite needs a $\sqrt{2}$ -fold speed boost to escape.

Using the approximation $g_h \approx g(1-2h/R)$ for large $h$ . This approximation works only when $h \ll R$ . For $h$ comparable to $R$ , use the exact formula $g_h = g(R/(R+h))^2$ . Using the approximation at $h = R$ gives $g_h = g(1-2) = -g$ , which is nonsensical.

Thinking astronauts in the ISS are beyond gravity. At ISS altitude (~400 km), $g \approx 8.7$ m/s $^2$ — only 11% less than on the surface. Astronauts feel weightless because they are in free fall (falling around Earth), not because gravity is absent.

Exam Weightage and Strategy

Gravitation carries 5-7 marks in CBSE Class 11 boards and 1-2 JEE Main questions per year. NEET tests it occasionally (1 question). The high-frequency PYQ topics: (1) variation of $g$ with height and depth, (2) escape velocity calculation, (3) satellite orbital velocity and energy, (4) Kepler’s third law. These four cover 90% of exam questions.

Memorise five formulas: $F = GMm/r^2$ , $g_h$ and $g_d$ formulas, $v_e = \sqrt{2gR}$ , $v_o = \sqrt{GM/r}$ , and $E = -GMm/2r$ . Practice 10 numericals covering different types. The chapter is formula-driven — once you have the formulas, the problem-solving is substitution.

Practice Questions

Q1. What is $g$ at height equal to Earth’s radius?

$g_h = g(R/(R+R))^2 = g/4 = 2.5$ m/s $^2$ (taking $g = 10$ ). At height $R$ , the distance from Earth’s centre is $2R$ , so gravity is $(1/2)^2 = 1/4$ of the surface value.

Q2. A satellite has period 2 hours. Find its orbital radius. ( $R = 6400$ km, $g = 10$ m/s $^2$ )

$T = 2\pi\sqrt{r^3/(gR^2)}$ . Rearranging: $r^3 = gR^2T^2/(4\pi^2)$ . $r^3 = 10 \times (6.4 \times 10^6)^2 \times (7200)^2/(4\pi^2)$ $= 10 \times 4.096 \times 10^{13} \times 5.184 \times 10^7 / 39.48$ $\approx 5.38 \times 10^{20}$ $r \approx 8.13 \times 10^6$ m = 8130 km. This is about 1730 km above the surface.

Q3. Find the ratio $v_e : v_o$ for a satellite near Earth’s surface.

$v_e/v_o = \sqrt{2gR}/\sqrt{gR} = \sqrt{2} \approx 1.414$ . Escape velocity is always $\sqrt{2}$ times the orbital velocity at the same radius.

Q4. At what depth is $g$ equal to $g$ at height $h$ (for small $h$ )?

$g(1-d/R) = g(1-2h/R)$ . So $d = 2h$ . You need to go twice as deep as the height to get the same reduction in $g$ . This is because $g$ decreases linearly with depth but quadratically (approximately) with height.

Q5. What is the total energy of a 200 kg satellite in orbit at 400 km above Earth? ( $g = 10$ m/s $^2$ , $R = 6400$ km)

$r = R + h = 6400 + 400 = 6800$ km $= 6.8 \times 10^6$ m. $E = -\frac{GMm}{2r} = -\frac{gR^2m}{2r} = -\frac{10 \times (6.4 \times 10^6)^2 \times 200}{2 \times 6.8 \times 10^6}$ $= -\frac{10 \times 4.096 \times 10^{13} \times 200}{1.36 \times 10^7} = -\frac{8.192 \times 10^{16}}{1.36 \times 10^7} \approx -6.02 \times 10^9$ J $\approx -6.02$ GJ.

The negative sign means the satellite is bound to Earth.

FAQs

Why is gravitational PE negative?

We define PE = 0 at infinity. To bring a mass from infinity to a point near another mass, gravity does positive work (it pulls the mass in, so the mass gains kinetic energy). By conservation of energy, PE must decrease — it goes from 0 to a negative value. The more negative the PE, the more tightly bound the system.

What is a geostationary orbit?

An orbit where the satellite has the same angular velocity as Earth — it stays fixed above one point on the equator. Requirements: circular orbit, equatorial plane, altitude ~36,000 km, period = 24 hours. Used for communication and weather satellites because ground antennas do not need to track the satellite.

Does gravity depend on the medium between two objects?

No. Gravitational force is independent of the medium — it cannot be shielded, blocked or amplified. Unlike electrostatic force (reduced by a dielectric) or magnetic force (affected by materials), gravity acts through all matter identically.

What are Kepler’s three laws?

Orbits are ellipses with the Sun at one focus. 2. The line from Sun to planet sweeps equal areas in equal times (conservation of angular momentum). 3. $T^2 \propto a^3$ — the square of the period is proportional to the cube of the semi-major axis.

What is the Roche limit?

The minimum distance at which a celestial body, held together only by its own gravity, can orbit a larger body without being torn apart by tidal forces. Inside the Roche limit, tidal forces exceed the body’s self-gravity. Saturn’s rings exist within its Roche limit — they are debris that could not coalesce into a moon.