Gravitation notes: CBSE Class 11 Physics

1. Kepler’s laws, the law of gravitation and superposition

Kepler’s laws: (1) each planet moves in an ellipse with the Sun at one focus; (2) the line from the Sun to the planet sweeps equal areas in equal times (angular momentum is conserved because gravity is central, so a planet is fastest at perihelion: $v_p r_p = v_a r_a$ ); (3) $T^2 \propto a^3$ with $a$ the semi-major axis. For a circular orbit $T^2 = \dfrac{4\pi^2}{GM}r^3$ . A body at $2.25$ AU has $T = 3.375$ years.
Newton’s law: $F = \dfrac{G m_1 m_2}{r^2}$ , attractive, along the line joining the bodies, equal and opposite on the two; $G = 6.67 \times 10^{-11}$ N m² kg⁻². The Sun pulls the Earth with $3.56 \times 10^{22}$ N ( $M_S = 2.0 \times 10^{30}$ kg, $r = 1.5 \times 10^{11}$ m).
Superposition: the force on a body from several masses is the vector sum of the individual pulls. A 4.0 kg mass with 2.0 kg at 0.30 m and 3.0 kg at 0.40 m (perpendicular) feels $7.76$ nN at $40.2^\circ$ from the line to the 2.0 kg mass.
A uniform sphere acts on outside bodies as if its whole mass were at its centre.

Equal areas in equal times: the planet speeds up near the Sun.

T² = a³ (T in years, a in AU)
six planets (a³, T²)

T² is proportional to a³: all planets lie on one straight line.

Add the pulls as vectors: gravity obeys superposition.

Table of Kepler three laws, what each states and its physical reason
Law	Statement	Physical reason
Orbits	A planet moves in an ellipse with the Sun at one focus	Inverse-square attraction
Areas	The line from Sun to planet sweeps equal areas in equal times	Force along the line: angular momentum is constant
Periods	T² is proportional to a³ (a = semi-major axis)	Gravity falls as 1/r²

Table of the mean distance in astronomical units and the period in years of six planets and the ratio T squared over a cubed, which is about one for each
Planet	a (AU)	T (year)	T² / a³
Mercury	0.387	0.241	1.00
Venus	0.723	0.615	1.00
Earth	1	1	1.00
Mars	1.524	1.881	1.00
Jupiter	5.203	11.86	0.999
Saturn	9.537	29.46	1.00

Formulas

Law of gravitation
$F = \dfrac{G m_1 m_2}{r^2}$
Law of periods, circular orbit
$T^2 = \dfrac{4\pi^2}{GM}\,r^3$
Law of areas at the extreme points
$v_p r_p = v_a r_a$

Key terms

Perihelion and aphelion: The nearest and farthest points of a planet’s orbit from the Sun; the speed is greatest at perihelion.
Semi-major axis: Half the longest diameter of the ellipse; the length that enters T² ∝ a³.
Superposition: The net gravitational force is the vector sum of the forces due to the individual masses.

2. Acceleration due to gravity and its variation

At the surface: $g = \dfrac{GM}{R^2} = \dfrac{4}{3}\pi G\rho R$ ; independent of the mass of the falling body. On the Moon $g = 1.62$ m s⁻², about one sixth of the Earth’s.
Above the surface: $g' = g\left(\dfrac{R}{R+h}\right)^2 \approx g\left(1 - \dfrac{2h}{R}\right)$ for $h \ll R$ . At $h = R$ , $g' = g/4$ . At 320 km the exact value is $8.889$ m s⁻² against $8.820$ from the approximation.
Below the surface (uniform Earth): only the inner sphere attracts, so $g' = g\left(1 - \dfrac{d}{R}\right)$ , falling linearly to zero at the centre; at 320 km deep $g' = 9.310$ m s⁻². Over the same small distance, g falls twice as fast above the surface (2h/R) as below it (d/R).
Weight $mg$ varies with place while mass does not; the Earth’s rotation and its non-uniform density are not considered here.

g falls both above and below the surface.

inside (uniform density): g′ = g r/R
outside: g′ = g R²/r²

g is zero at the centre, largest at the surface and falls as 1/r² outside.

Table of the acceleration due to gravity at the surface, at a height, at a depth and at the centre of the Earth
Where	g′	Condition
At the surface	G M / R²	Spherical Earth of mass M
Height h above	g R² / (R + h)² ≈ g (1 − 2h/R)	h much less than R for the approximation
Depth d below	g (1 − d/R)	Uniform density; only the inner sphere pulls
Centre	0	Equal pulls all round

Formulas

g at the surface
$g = \dfrac{GM}{R^2}$
g at height h
$g' = g\left(1 + \dfrac{h}{R}\right)^{-2}$
g at depth d
$g' = g\left(1 - \dfrac{d}{R}\right)$

Key terms

Acceleration due to gravity: The acceleration of a freely falling body, GM/R² at the surface of a planet.
Inner sphere: At depth d only the mass within radius R − d attracts a body; the shell above exerts no net force.

3. Gravitational potential energy and potential

With zero at infinity, $U(r) = -\dfrac{GMm}{r}$ : negative, rising towards zero as $r$ increases. A $250$ kg satellite at $1.0 \times 10^7$ m has $U = -1.00 \times 10^{10}$ J.
Potential $V = U/m = -\dfrac{GM}{r}$ is a scalar, so potentials of several masses add as numbers; $U = mV$ . Two masses (5.0 kg and 10 kg) 0.50 m apart give $V = -3.89 \times 10^{-9}$ J kg⁻¹ at 0.20 m from the lighter one.
Gravity is conservative: the work between two points depends only on the end points.
Near the surface: $\Delta U = \dfrac{mgRh}{R+h} \approx mgh$ for $h \ll R$ ; the exact value is smaller than $mgh$ because g decreases with height. Lifting 120 kg through 3200 km needs $2.51 \times 10^{9}$ J, not $3.76 \times 10^{9}$ J.

U = −G M m / r

U is negative everywhere and rises towards 0 as r increases.

Table comparing gravitational potential energy and gravitational potential: symbol, unit and kind
Quantity	Symbol	Unit	Kind
Gravitational potential energy of m at distance r	U = −G M m / r	joule	Scalar, belongs to the pair
Gravitational potential at distance r	V = −G M / r	joule per kilogram	Scalar, belongs to the point
Link	U = m V		Potential energy = mass × potential

Table comparing the exact change of potential energy per kilogram with g h for heights from 1 kilometre to 6400 kilometres
Height h	Exact ΔU per kg (kJ)	g h per kg (kJ)	g h is too big by
1 km	9.80	9.80	0.0156 %
10 km	97.8	98.0	0.156 %
100 km	965	980	1.54 %
640 km	5700	6270	9.09 %
6400 km	31400	62700	50.0 %

Formulas

Potential energy
$U = -\dfrac{GMm}{r}$
Potential
$V = -\dfrac{GM}{r}$
Raising a body from the surface to height h
$\Delta U = \dfrac{mgRh}{R+h}$

Key terms

Gravitational potential energy: The work done by an external agent in bringing a mass from infinity to a point; negative, −GMm/r.
Gravitational potential: Potential energy per unit mass, −GM/r, a scalar.

4. Escape speed, orbital speed and energy of a satellite

Escape speed: $\tfrac{1}{2}mv_e^2 - \dfrac{GMm}{R} = 0$ gives $v_e = \sqrt{\dfrac{2GM}{R}} = \sqrt{2gR}$ : 11.2 km s⁻¹ for the Earth, 2.37 km s⁻¹ for the Moon, 5.01 km s⁻¹ for Mars. It does not depend on the projectile’s mass or the direction of launch.
Circular orbit: $\dfrac{mv^2}{r} = \dfrac{GMm}{r^2}$ gives $v = \sqrt{\dfrac{GM}{r}}$ and $T = 2\pi\sqrt{\dfrac{r^3}{GM}}$ , independent of the satellite’s mass; $v_e = \sqrt{2}\,v$ . Close to the Earth $v = \sqrt{gR} \approx 7.9$ km s⁻¹ and $T \approx 85$ minutes. At $h = R$ : $v = 5.60$ km s⁻¹, $T \approx 3.99$ h.
Energy: $K = \dfrac{GMm}{2r}$ , $U = -\dfrac{GMm}{r}$ , $E = -\dfrac{GMm}{2r} = -K$ . Negative E means the satellite is bound; $|E|$ is the binding energy. Raising a 400 kg satellite from $8.0 \times 10^6$ m to $1.2 \times 10^7$ m needs $3.33 \times 10^{9}$ J although its speed decreases.
Air drag lowers E, so the orbit shrinks and the speed rises.
Geostationary and polar satellites and weightlessness are not treated in this chapter (see the README for the scope decision).

Faster launches go farther; at root two times the circular speed the body escapes.

Force towards the centre, velocity along the tangent.

orbital speed v = √(GM/r)
escape speed = √2 × orbital speed

Higher orbits are slower; escape speed is always √2 times the orbital speed at the same r.

K = G M m / 2r
U = −G M m / r
E = K + U = −G M m / 2r

E = −K = U/2 at every orbit radius.

Table of the mass, radius and escape speed of the Earth, the Moon and Mars computed from root of 2GM over R
Body	M (kg)	R (m)	Escape speed
Earth	6.0 × 10²⁴	6.4 × 10⁶	11.2 km/s
Moon	7.35 × 10²²	1.74 × 10⁶	2.37 km/s
Mars	6.4 × 10²³	3.4 × 10⁶	5.01 km/s

Table of formulae for a satellite in a circular orbit: speed, period, kinetic, potential and total energy, and the escape speed
Quantity (circular orbit, radius r)	Formula
Orbital speed	v = √(G M / r)
Period	T = 2π √(r³ / G M)
Kinetic energy	K = G M m / 2r
Potential energy	U = −G M m / r
Total energy	E = −G M m / 2r = −K
Escape speed at the same r	√2 × v

Formulas

Escape speed
$v_e = \sqrt{\dfrac{2GM}{R}}$
Orbital speed
$v = \sqrt{\dfrac{GM}{r}}$
Total energy of a circular orbit
$E = -\dfrac{GMm}{2r}$

Key terms

Escape speed: The least launch speed that lets a body reach infinity with no speed left; √(2GM/R).
Orbital speed: The speed of a satellite in a circular orbit, √(GM/r).
Binding energy: The magnitude of the total energy of a bound satellite; the energy needed to free it.

Exam tips

Quote Kepler’s laws in words; for T² ∝ a³ use the semi-major axis a of an ellipse.
Add forces of several masses as vectors; potentials as numbers.
g′ falls as the inverse square above the surface (exact) and linearly below it (uniform Earth).
Say that the potential energy is zero at infinity before using −GMm/r; bound energies are negative.
Derive v_e from energy and v_orbit from force; use gR² for GM when g and R are given.

1. Kepler’s laws, the law of gravitation and superposition

Formulas

Key terms

2. Acceleration due to gravity and its variation

Formulas

Key terms

3. Gravitational potential energy and potential

Formulas

Key terms

4. Escape speed, orbital speed and energy of a satellite

Formulas

Key terms

Exam tips

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