Class XI  Chapter 07

 
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Class XI · Chapter 07

Gravitation

The force that keeps your feet on the ground and the Moon in orbit is the exact same force — one law from Newton connects falling apples to entire planetary systems.

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1Kepler's Laws of Planetary Motion

Long before Newton explained why, Johannes Kepler worked out precisely how planets move, purely from careful observation.

  1. Law of Orbits: every planet moves in an ellipse, with the Sun at one focus (not the centre).
  2. Law of Areas: the line joining a planet to the Sun sweeps out equal areas in equal time intervals — meaning a planet moves faster when closer to the Sun, slower when farther away.
  3. Law of Periods: the square of a planet's orbital period is proportional to the cube of its orbit's semi-major axis: T² ∝ a³.
A Hidden Connection The law of areas isn't a separate rule — it's exactly the conservation of angular momentum (Chapter 6) in disguise, since gravity always points directly at the Sun and exerts no torque about it.

2Newton's Universal Law of Gravitation

Newton's breakthrough was recognising that the force pulling an apple down and the force holding the Moon in orbit are the exact same force, following one universal rule.

Universal Law of Gravitation
F = G m₁m₂ / r²
Every pair of masses attracts every other pair — always attractive, always along the line joining them, and following an inverse-square law exactly like Coulomb's Law (Chapter 1 of Class XII) does for charge.

3The Gravitational Constant

Gravitational Constant
G ≈ 6.674 × 10⁻¹¹ N·m²/kg²
G is famously tiny, which is exactly why gravity between everyday objects is imperceptible — it only becomes significant when at least one mass is astronomically large.

4Acceleration Due to Gravity on Earth

Applying the universal law to a mass at Earth's surface, with Earth treated as a uniform sphere, gives the familiar constant g.

Surface Gravity
g = GM / R²
M, R = Earth's mass and radius. This is exactly why g ≈ 9.8 m/s² is the same for a feather and a bullet — mass cancels out entirely (F = mg = GMm/R² → g = GM/R², independent of the falling object's own mass).

5Variation of g with Height and Depth

Above the Surface (height h)
gh = g (1 − 2h/R), for h ≪ R
g decreases as you go higher — less mass "pulls" from farther away, and you're farther from the centre.
Below the Surface (depth d)
gd = g (1 − d/R)
g also decreases going down — only the mass enclosed within your current radius contributes to gravity felt there; g reaches exactly zero at Earth's centre.

6Gravitational Potential Energy

Chapter 5's U = mgh only works for small heights near Earth's surface, where g is essentially constant. For larger distances, gravitational potential energy needs the full inverse-square-consistent form:

Gravitational Potential Energy
U = − G M m / r
Taken as zero at infinite separation — hence the negative sign everywhere closer, reflecting that gravity is attractive and energy must be added to pull two masses apart.

7Escape Speed

Escape speed is the minimum launch speed needed for an object to break entirely free of a body's gravity, reaching infinity with exactly zero leftover kinetic energy.

Escape Speed
ve = √(2GM/R) = √(2gR)
For Earth, this works out to about 11.2 km/s — notice it doesn't depend on the launched object's mass or launch direction at all, only on the planet's own mass and radius.

8Earth Satellites

A satellite in a stable circular orbit is in continuous free fall, with gravity supplying exactly the centripetal force needed to keep it curving around the Earth rather than flying off straight.

Orbital Velocity
vo = √(GM/r)
r = orbital radius (measured from Earth's centre, not altitude). Lower orbits need higher speed to maintain the tighter curve.
Orbital Period
T² = 4π² r³ / GM
This is literally Kepler's third law, now derived from first principles rather than just observed.
Total Energy of an Orbiting Satellite
E = − GMm / 2r
Negative total energy is the signature of a bound orbit — the satellite doesn't have enough energy to escape to infinity, which is exactly what keeps it circling rather than flying away.

9Geostationary and Polar Satellites

  • Geostationary satellites: orbit directly above the equator with a period of exactly 24 hours, matching Earth's own rotation — so they appear fixed at one point in the sky, at an altitude of roughly 36,000 km. Ideal for communication and broadcast satellites, since a receiving dish never needs to track them.
  • Polar satellites: orbit at much lower altitude, passing over (or near) both poles on each revolution, while Earth rotates beneath them — over time, this sweeps the satellite's view across the entire planet, making them ideal for mapping, weather monitoring, and surveillance.

10Weightlessness

Astronauts orbiting Earth aren't weightless because gravity has vanished — at typical orbital altitudes, gravity is only slightly weaker than at the surface. They feel weightless because they and their spacecraft are both in continuous free fall together, falling around the Earth at exactly the same rate — with nothing pressing them against any surface to register as "weight," the same sensation as the brief drop of a fast elevator, sustained indefinitely.

Formula Summary

Universal Gravitation
F = Gm₁m₂/r²
Surface Gravity
g = GM/R²
g at Height h
g_h = g(1−2h/R)
g at Depth d
g_d = g(1−d/R)
Gravitational PE
U = −GMm/r
Escape Speed
v_e = √(2GM/R)
Orbital Velocity
v_o = √(GM/r)
Orbital Period
T² = 4π²r³/GM
Satellite Total Energy
E = −GMm/2r

Solved Examples

Example 1 · Escape Speed from Earth

Find the escape speed from Earth's surface. (M = 5.97×10²⁴ kg, R = 6.37×10⁶ m, G = 6.674×10⁻¹¹ N·m²/kg²)

Solution: ve = √(2GM/R) = √[(2 × 6.674×10⁻¹¹ × 5.97×10²⁴) / (6.37×10⁶)].

= √(1.25×10⁸) ≈ 1.12×10⁴ m/s ≈ 11.2 km/s — the well-known figure for launching anything permanently away from Earth.

Example 2 · Orbital Velocity

Find the orbital speed of a satellite 300 km above Earth's surface. (Use R = 6370 km, M = 5.97×10²⁴ kg)

Solution: r = R + h = 6370 + 300 = 6670 km = 6.67×10⁶ m.

vo = √(GM/r) = √[(6.674×10⁻¹¹ × 5.97×10²⁴) / (6.67×10⁶)] ≈ √(5.97×10⁷) ≈ 7.73×10³ m/s ≈ 7.73 km/s.

Quick Check

1. Kepler's second law (equal areas in equal times) is a direct consequence of conservation of:
Energy
✓ Angular momentum (correct)
Linear momentum
Mass
2. Astronauts in orbit feel weightless because:
Gravity is zero at orbital altitude
✓ They and their spacecraft are in continuous free fall together (correct)
Their mass becomes zero in orbit
Air resistance cancels gravity
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System of Particles and Rotational Motion

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Mechanical Properties of Solids

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