Gravitation: theory | Physics Circle

CONCEPT

Newton’s Law of Gravitation

In one line
Every body attracts every other body with a force proportional to the product of their masses and inversely proportional to the square of the distance between them.

THE LAW

8.1.1 Newton’s law of gravitation

Equal and opposite forces along the line joining the centres

F = Gm₁m₂r²

The force acts along the line joining the two particles.
The two forces are an action–reaction pair: the earth pulls the moon as hard as the moon pulls the earth.

THE CONSTANT G

8.1.2 Universal gravitational constant

G = 6.67 × 10⁻¹¹ N m² kg⁻²

G equals the force between two unit masses kept unit distance apart.
Dimensions: [M⁻¹L³T⁻²].
First measured by Cavendish with a torsion balance.
G does not depend on the nature or size of the bodies, or on the medium.

PROPERTIES

8.1.3 Nature of the gravitational force

Always attractive.
Independent of the medium between the bodies.
Acts over a wide range: planets to atoms.
Central: along the line joining the centres.
Weakest force in nature: F_nuclear > F_{electromagnetic} > F_{gravitational}.
Between two electrons, gravity is about 10⁻⁴³ of the electrostatic force.
Conservative: work around a closed path is zero.

SUPERPOSITION

8.1.4 Force due to many particles

The force between two particles is not affected by other particles.
The net force is the vector sum.

F = F₁ + F₂ + F₃ + …

Equal masses placed symmetrically around a point give zero force there.
A particle at the centre of a uniform ring feels zero force.
Square of side a with mass m at each corner: force on one mass is

F = (2√2 + 12) Gm²a²

FINITE BODIES

8.1.5 What distance to use

Irregular bodies have no single r; for spheres take the distance between centres

The law is stated for point masses.
A uniform sphere acts as a point mass for any point outside it.
Two identical spheres in contact are 2R apart, with m = 43πR³ρ:

F ∝ R⁴ρ²

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Vector form of the law

F₁₂ = −Gm₁m₂r² r₂₁ = −F₂₁

r₂₁ is the unit vector from m₂ to m₁; the minus sign shows the force on m₁ points towards m₂.
F₁₂ = −F₂₁ is Newton’s third law.

Sphere with a cavity

Treat the cavity as a filled sphere plus a sphere of negative mass.
Force of the rest = force of the full sphere − force of the removed part.
A part of radius Rn has mass Mn³. Measure its distance from its own centre.

Two bodies released from rest

Their centre of mass stays at rest, so distance covered ∝ one over the mass.
They collide when the centres are R₁ + R₂ apart.

Continuous bodies

Semicircular wire of length L, mass M, at the centre: F = 2πGMmL².
Rod with λ(x): F = Gm ∫ λ dxx².

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CONCEPT

Acceleration Due to Gravity

In one line
Near a planet every body falls with the same acceleration g, whatever its mass. It depends only on the planet’s mass and radius.

g FROM THE LAW

8.2.1 Acceleration due to gravity

Pull on a body at the surface

Pull of the earth on the body = mg.

mg = GMmR²

g = GMR²

Put M = volume × density = 43πR³ρ:

g = 43πρGR

g does not depend on the mass, shape or density of the falling body.
g is a vector, directed towards the centre of the planet.
Dimensions [LT⁻²]; average value 9.8 m s⁻².
GM = gR²: use it to replace GM in every formula.

COMPARING PLANETS

8.2.2 Which formula to use

Given	Use
M and R	g ∝ MR²
ρ and R	g ∝ ρR
same density	g ∝ R, and g ∝ M^1/3
same mass	g ∝ 1R²
same g	ρ ∝ 1R

Radius falls by x% (mass fixed): g rises by 2x%.
Mass rises by x% (radius fixed): g rises by x%.
Density falls by x% (radius fixed): g falls by x%.
Diameter halved (mass fixed): g becomes 4g.

USES OF g

8.2.3 What changes with g on a planet

Weight: W = mg, while the mass m stays the same.
Work in lifting a body: W = mgh.

pendulum: T = 2π√lg ⇒ T ∝ 1√g

A seconds pendulum keeps T = 2 s, so its length l ∝ g.

height of a jump: h = v²2g ⇒ h ∝ 1g

MASS AND DENSITY

8.2.4 Weighing the earth

M = gR²G

With g = 9.8 m s⁻² and R = 6400 km, M ≈ 6 × 10²⁴ kg.

ρ = 3g4πGR

ρ ≈ 5478 kg m⁻³ for the earth.
ρ is directly proportional to g.

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Inertial and gravitational mass

m_i = Fa · m_g = FI

Inertial mass measures opposition to acceleration; an inertial balance measures it.
Gravitational mass measures the gravitational pull; a spring balance measures it.
Both are scalars with the same unit, and their ratio is 1.
A moving body: m = m₀√1 − v²/c².

Mass and weight

Mass	Weight
quantity of matter	pull of the earth on the body
does not change with g	changes with g
never zero	zero at the centre of the earth
kg, [M], scalar	N, [MLT⁻²], vector

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CONCEPT

Variation of g

In one line
g is largest at the surface of the earth. It falls as you go up or down, and it rises from the equator to the poles.

HEIGHT

8.3.1 g above the surface

r = R + h

g′ = g (RR + h)² = gR²r²

At infinity g′ = 0.
At h = R: g′ = g4. At h = 2R: g′ = g9.
Weight changes in the same way: W′ = mg′.

When h ≪ R:

g′ = g(1 − 2hR)

% decrease = 2hR × 100

Pendulum at height h: T′T = R + hR.
Seconds pendulum: l_hl_s = (RR + h)².

DEPTH

8.3.2 g below the surface

Only the inner sphere pulls the body

g′ = g(1 − dR)

% decrease = dR × 100

At the centre d = R: g′ = 0, so weight is zero.
At d = R2: g′ = g2.
To get g′ = gn: d = R(n − 1n).

Frequency of a pendulum: n′ = n √(1 − dR).
Capillary rise h ∝ 1g: h₂h₁ = RR − d in a mine.

HEIGHT AND DEPTH

8.3.3 Comparing the two

g rises in a straight line up to R, then falls

Same g at height h and depth d (both small): d = 2h, not d = h.
Going up, g falls twice as fast as going down.
Ratio: g_dg_h = R − dR − 2h.

SHAPE

8.3.4 Equator and poles

The equatorial radius is about 21 km longer

g = GMR² ⇒ g_eg_p = R_p²R_e²

R_equator > R_pole, so g_pole > g_equator.
Due to shape alone: g_p = g_e + 0.018 m s⁻².

ROTATION

8.3.5 g at latitude λ

Point P moves in a circle of radius r = R cos λ

The latitude λ is the angle from the equatorial plane.

g′ = g − ω²R cos²λ

Place	λ	g′
pole	90°	g (no effect)
equator	0°	g − ω²R (largest effect)

g_p − g_e = ω²R = 0.034 m s⁻² due to rotation.
Shape and rotation together: g_p = g_e + 0.052 m s⁻².
If the earth stops rotating, g at the equator rises by ω²R.
A body hung by a string at the equator:

T = GMmR² − mω²R

WEIGHTLESS AT THE EQUATOR

8.3.6 How fast must the earth spin?

ω = √gR = 1800 rad s⁻¹

T = 2π√(Rg) ≈ 84.6 minutes ≈ 1.4 h.
That is about 17 times faster than now.
To weigh x times the present weight:

ω = √(1 − x) gR

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Tunnel through the earth

Inside the earth the pull is proportional to the distance from the centre.
A particle in a tunnel along the diameter therefore moves in SHM.

T = 2π√Rg ≈ 84.6 minutes

Spring balance: equator and height

Spring balance reading at the equator = reading at height h over the pole.

h = ω²R²2g

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CONCEPT

Gravitational Field

In one line
The space around a mass where its pull can be felt is its gravitational field. Its strength at a point is the force on a unit mass placed there.

INTENSITY

8.4.1 Gravitational field intensity

I = Fm

A vector, directed towards the source of the field.
Unit: N/kg or m/s². Dimensions: [M⁰LT⁻²].

Field at distance r from a point mass M

point mass: I = GMr²

At the surface of the earth, I = g.
I = 0 at infinity.
Field lines of an isolated mass point radially inwards.

Field lines around an isolated mass

Several masses: add the fields as vectors.

I_net = I₁ + I₂ + I₃ + …

NULL POINT

8.4.2 Where the field is zero

P lies between the masses, nearer the smaller one

Gm₁x² = Gm₂(d − x)²

x = √m₁ d√m₁ + √m₂

Earth and moon (81 : 1): x = 9d10 from the earth.
A test mass at P feels no force.

FIELD FROM POTENTIAL

8.4.3 Link with V

I = −dVdr

V = −Kx gives I = −Kx².

SOLID SPHERE

8.4.4 Uniform sphere of mass M

Field grows linearly inside and falls outside

Outside, r > R	Surface, r = R	Inside, r < R
GMr²	GMR²	GMrR³

Point inside and point outside: I_outI_in = R³r₁²r₂.
Largest at the surface; zero at the centre.

SPHERICAL SHELL

8.4.5 Thin shell of mass M

Zero inside, then a jump at the surface

Outside, r > R	Surface, r = R	Inside, r < R
GMr²	GMR²	0

Only the mass inside the point counts.
Sphere inside a shell: between them only the sphere pulls; outside the shell both pull.

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Uniform circular ring

Point P on the axis of a ring of mass M

on the axis: I = GMr(a² + r²)^3/2 · centre: I = 0

I is largest at r = a√2.

Uniform disc

Point P on the axis of a disc of mass M

on the axis: I = 2GMa²(1 − cos θ) · centre: I = 0

Sphere with non-uniform density

Inside, use the mass enclosed: I = G m(r)r², with m(r) = ∫ 4πr²ρ dr.
For maximum field set dIdr = 0.

Flux of the gravitational field

Any closed surface around mass M has flux of magnitude 4πGM.
Its size and shape do not matter.

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CONCEPT

Gravitational Potential & Energy

In one line
Potential is the work per unit mass to bring a body from infinity to a point. It is a scalar and is negative everywhere.

POTENTIAL

8.5.1 Gravitational potential

V = Wm = −∫ I dr

point mass: V = −GMr

Scalar. Unit: J/kg. Dimensions: [M⁰L²T⁻²].
Reference: V = 0 at infinity.
Given field I(r), find V by integrating: dV = −I dr.
Uniform field: V = −(E_xx + E_yy).

Moving unit mass from A to B

V_B − V_A = GM(1r_A − 1r_B)

MANY MASSES

8.5.2 Scalar addition

Potentials simply add at P

V = −G(M₁r₁ + M₂r₂ + M₃r₃ + …)

Potential at a null point is not zero.
Masses m and M, distance d, at the null point:

V = −Gd(√m + √M)²

SHELL

8.5.3 Thin spherical shell

Constant inside

Outside	Surface	Inside
−GMr	−GMR	−GMR

Inside a shell the field is zero, the potential is constant but not zero.

SOLID SPHERE

8.5.4 Uniform sphere

Deepest at the centre

Outside	Surface	Inside
−GMr	−GMR	−GM2R³(3R² − r²)

V_centre = −3GM2R = 32 V_surface

POTENTIAL ENERGY

8.5.5 Energy of position

Work done bringing m from infinity to r

U = mV = −GMmr

Negative, because the force is attractive.
Zero (its maximum) at infinity.
It rises (becomes less negative) as r increases.
Conservative: work depends only on the end points, not the path.
At the surface: U = −GMmR.

WORK AGAINST GRAVITY

8.5.6 Raising a body by height h

ΔU = GMm(1R − 1R + h) = GMmhR(R + h)

ΔU = mgh1 + hR

h	ΔU
R	12 mgR
2R	23 mgR
3R	34 mgR
nR	nn + 1 mgR
h ≪ R	mgh

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Potential of a ring

on the axis: V = −GM√a² + r² · centre: V = −GMa

Energy of a system of masses

U = −G(m₁m₂r₁₂ + m₂m₃r₂₃ + m₁m₃r₁₃ + …)

Count each pair once.
n masses have n(n − 1)2 pairs.
Work to expand or shrink a system = U_final − U_initial.

Energy to break up the earth

U_sphere = −3GM²5R

Supplying this much energy takes all its mass to infinity.

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CONCEPT

Escape Velocity

In one line
Escape velocity is the least speed a body needs at the surface to leave the planet’s pull for ever. It does not depend on the body’s mass or direction.

THE FORMULA

8.6.1 Kinetic energy pays for the escape

Work to go from the surface to infinity:

W = GMmR

12 mv_e² = GMmR

v_e = √2GMR = √2gR = R√8πGρ3

For the earth: 11.2 km s⁻¹.
Independent of the mass of the body and of the angle of projection.
Escape energy = GMmR.
Ant or elephant: the same speed.

COMPARING PLANETS

8.6.2 Which formula to use

Given	Use
M and R	v_e ∝ √MR
g and R	v_e ∝ √gR
same density	v_e ∝ R
ρ and R	v_e ∝ R√ρ

Radius doubled at the same density: escape velocity doubled.
Radius doubled at the same mass: it becomes 1 ÷ √2 times.

v_e = √2 v_o

v_o = orbital speed close to the surface.

v LESS THAN v_e

8.6.3 Maximum height reached

12mv² − GMmR = −GMmR + h

h = Rv²v_e² − v² = v²R2gR − v²

If v = kv_e:

h = Rk²1 − k²

v	h
12 v_e	R3
13 v_e	R8
KE = half the escape KE	R

v GREATER THAN v_e

8.6.4 Speed far away

v′ = √v² − v_e²

v = 3v_e gives v′ = 2√2 v_e.

FALLING BODIES

8.6.5 Speed on reaching the surface

The same energy equation, read backwards.

from distance R₀: v = √2GM(1R − 1R₀)

from height h: v = √2gRhR + h

From infinity: v = v_e = 11.2 km s⁻¹.
Small h: v = √(2gh).
KE gained between two points: GMm(1r₂ − 1r₁).

OTHER STARTING POINTS

8.6.6 Not from the surface

Use the potential at the starting point:

12 v² = |V|

at distance r: v = √2GMr

From height h = 3R, r = 4R: v = √GM2R.
Between earth and moon: V has a term from each.
At a distance r3 from M₁, on the line to M₂:

v = √6Gr(M₁ + M₂2)

A satellite near the surface needs its speed raised by the factor √2 (41.4% more) to escape.

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Why some planets have no atmosphere

v_rms = √3RTM

A planet keeps its atmosphere if v_rms of its gas molecules is less than its escape velocity.
Earth keeps it. The moon has v_rms > v_e, so it has no atmosphere.

Black hole

If the escape velocity equals the speed of light c, nothing can escape, not even light.

R = 2GMc²

Two planets starting at rest far apart

When they are a distance d apart, use momentum and energy conservation together.
Two planets: v₁ = m₂√2Gd(m₁ + m₂).

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CONCEPT

Kepler’s Laws of Planetary Motion

In one line
Planets move in ellipses with the sun at a focus, sweep equal areas in equal times, and obey T² ∝ a³.

THE THREE LAWS

8.7.1 Kepler’s laws

Law of orbits: every planet moves in an ellipse with the sun at one focus.
Law of areas: the line from the sun to a planet sweeps equal areas in equal times.
Law of periods: T² ∝ a³, where a is the semi-major axis.

The sun is at one focus

a = r₁ + r₂2

LAW OF AREAS

8.7.2 Areal velocity

The two shaded areas take equal times

dAdt = rv2 = L2m

Comes from conservation of angular momentum.
The planet moves fastest when nearest the sun.
Time taken ∝ area swept.

LAW OF PERIODS

8.7.3 Comparing planets

(T₂T₁)² = (r₂r₁)³

For the earth T = 1 year.
Period 8 times ⇒ radius 4 times.
Radius 14 ⇒ period 18.
Radius 3 times ⇒ period 3√3 times.
These laws hold for satellites too.

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Perigee and apogee

Force is central ⇒ no torque ⇒ angular momentum conserved.

m v_p r_p = m v_a r_a

v_pv_a = r_ar_p = 1 + e1 − e

r_p = a − c and r_a = a + c, where e = c ÷ a.

Speeds in terms of eccentricity

v_p = √GMa × 1 + e1 − e

v_a = √GMa × 1 − e1 + e

Kinetic energy is greatest at perigee and least at apogee.
Total energy stays constant along the orbit.

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CONCEPT

Satellites: Orbital Velocity & Period

In one line
Gravity supplies the centripetal force of a satellite. So its speed and period depend on the orbit radius and the planet’s mass, not on the satellite’s mass.

ORBITAL VELOCITY

8.8.1 Speed in a circular orbit

r = R + h

mv²r = GMmr²

v = √GMr = R√gR + h

Independent of the mass of the satellite.
Depends on the central mass and the orbit radius.

v ∝ 1√r

Planets around the sun obey the same rule.
r four times ⇒ speed half.
If gravity vanished, the satellite would fly off along the tangent.
A ball released from a satellite stays in the same orbit.

NEAR THE SURFACE

8.8.2 h ≪ R

v = √GMR = √gR ≈ 7.9 km s⁻¹

At radius nR: speed = 7.9√n km s⁻¹.
At altitude nR: speed = 7.9√(1 + n) km s⁻¹.

WHAT THE SPEED DOES

8.8.3 Horizontal velocity V_H at a height

V_H	Path
less than V_c	falls back to the earth
equal to V_c	circular orbit
between V_c and V_e	elliptical orbit
V_e or more	escapes

V_c is the critical (orbital) velocity at that height.

TIME PERIOD

8.8.4 One revolution

T = 2πrv = 2π√r³GM = 2π√(R + h)³gR²

T ∝ r^3/2

Independent of the mass of the satellite.
r four times ⇒ T eight times.
At h = R: T = 4π√(2Rg).

NEAR THE SURFACE

8.8.5 Shortest period

T = 2π√Rg ≈ 84.6 min ≈ 1.4 h

T = √3πGρ

T²ρ = 3πG, a constant.

OTHER FORCE LAWS

8.8.6 If F ∝ 1 ÷ rⁿ

Equate mv²/r to the force, then read off the powers.

v ∝ 1r^{(n − 1)/2} · T ∝ r^{(n + 1)/2}

n = 2 gives back T ∝ r^3/2.
n = 52 gives T ∝ r^7/4.

MUTUAL ORBITS

8.8.7 Particles moving under their own gravity

Find the net pull towards the centre; set it equal to mv²/r.
Two particles sit a diameter apart: the distance in the law is 2r, the radius in the orbit is r.

System	Result
two masses m, circle r	v = √Gm4r
three masses, side L	T ∝ L^3/2
four masses M, circle R	v = 12√GMR(1 + 2√2)

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Binary stars

Both stars move about the common centre of mass with the same ω and T.
For masses m₁, m₂ a distance d apart:

T = 2π√d³G(m₁ + m₂)

Other mass distributions

Find M enclosed within r, then mv²r = F.
Density ρ ∝ 1 ÷ r²: T ∝ r.
Long cylinder: field ∝ 1 ÷ r, so v is constant and T ∝ r.

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CONCEPT

Geostationary Satellite & Angular Momentum

In one line
A geostationary satellite has the same period as the earth’s spin, so it seems fixed in the sky. A satellite’s angular momentum grows as √r.

HEIGHT OF A SATELLITE

8.9.1 From its period

T² = 4π²(R + h)³gR²

h = (T²gR²4π²)^1/3 − R

GEOSTATIONARY SATELLITE

8.9.2 Fixed over one place

A communication satellite in the parking orbit

Orbit must be concentric and coplanar with the equator.
Sense of rotation same as the earth: west to east.
Period equal to the earth’s: T = 24 h.
The orbit is called the parking orbit.

ITS ORBIT

8.9.3 Radius, height and speed

ω²r = gR²r²

r = (gR²ω²)^1/3

r ≈ 42000 km ≈ 7R.
Height h ≈ 36000 km ≈ 6R.
Speed v ≈ 3.08 km s⁻¹.
Rearranged: R² = r³ω² ÷ g.

USING T ∝ r^3/2

8.9.4 Another satellite of the same planet

Add R to each height first: use the radius from the centre.

T₂24 h = (r₂r₁)^3/2

Radius half of the geostationary radius: T = 24 ÷ 2√2 = 6√2 h.
Radius one-fourth: T = 3 h.
Another planet: r³ ∝ MT².

ANGULAR MOMENTUM

8.9.5 Of a satellite

L = mvr = m√GMr

L ∝ √r

Depends on both masses and on the orbit radius.
Force is central ⇒ torque = 0 ⇒ L is conserved.
“Increased by eight times” means r′ = 9r, so L′ = 3L.

areal velocity dAdt = L2m

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Satellite seen from the earth

A satellite in the equatorial plane moving west to east has relative angular velocity ω_S − ω_E.

time between two overhead passes = T_ST_ET_E − T_S

If ω_S = ω_E the time is infinite: the satellite looks stationary.

Practise PYQs on this topicSatellites - Orbital Velocity & Period · 80 questions with solutions

CONCEPT

Satellite Energy & Weightlessness

In one line
A satellite’s total energy is negative: it is bound to the planet. Its kinetic energy is half the size of its potential energy.

THE THREE ENERGIES

8.10.1 In a circular orbit

U = −GMmr

K = 12mv² = GMm2r

E = U + K = −GMm2r

K = −E; U = 2E; U = −2K.
K : |U| = 1 : 2.
With GM = gR²: K = mgR²2r.
Orbit radius r = 4R: K = mgR8.
E ∝ mr: two satellites compare through m and r.

K is positive; U and E are negative

BINDING ENERGY

8.10.2 To free the satellite

BE = −E = GMm2r

The energy needed to take the satellite from its orbit to infinity.

LAUNCHING

8.10.3 From the surface into orbit

energy = E_orbit − E_surface = GMmR − GMm2r

Altitude	r	Energy
R	2R	3GMm4R
2R	3R	5GMm6R
3R	4R	7GMm8R

Add R to the altitude before using the formula.

RAISE AND ORBIT

8.10.4 Two different jobs

to raise to h: E₁ = GMmhR(R + h)

to give orbital speed: E₂ = GMm2(R + h)

E₁E₂ = 2hR

CHANGE OF ORBIT

8.10.5 Moving to a higher orbit

From r₁ to r₂

W = GMm2(1r₁ − 1r₂)

Quantity	As r rises	Relation
speed	falls	v ∝ 1√r
period	rises	T ∝ r^3/2
linear momentum	falls	p ∝ 1√r
angular momentum	rises	L ∝ √r
kinetic energy	falls	K ∝ 1r
potential energy	rises	U ∝ −1r
total energy	rises	E ∝ −1r
binding energy	falls	BE ∝ 1r

WEIGHTLESSNESS

8.10.6 Weight is the reaction of the support

You feel your weight only when something opposes it.
If the support gives no reaction, the apparent weight is zero.

Gravity alone provides the acceleration

GMmr² − N = mGMr² ⇒ N = 0

Seen in free fall, in an orbiting satellite and at null points in space.
Gravity is not zero in a satellite; it is what keeps the satellite in orbit.
A pendulum in a satellite does not oscillate: T is infinite.
A spring balance still works, by measuring a force other than weight.
On the moon a person is not weightless, because the moon’s own gravity acts.

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Energy in an elliptical orbit

KE is largest at perigee, PE is largest at apogee

E = −GMm2a

a is the semi-major axis; E depends only on a.

Collisions in orbit

Find the speed just after the collision by conserving momentum.
Same speed as orbital speed: circular. Different but less than escape speed: ellipse.
A satellite whose speed is raised by the factor √2 escapes.

Splitting and launching

Rockets ejected: use momentum conservation for the parts.
Energy accounts must include both KE and PE of every part.

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EXAM-DAY CARD

Just see this before the exam

F = Gm₁m₂r²

Newton

g = GMR²

at the surface

I = GMr²

field

v_e = √2gR

escape

T² ∝ a³

Kepler

E = −GMm2r

satellite

FORMULA SHEET

Whole chapter on one page

Law & g

F = Gm₁m₂r²

g = GMR² = 43πρGR

M = gR²G, ρ = 3g4πGR

Variation of g

g′ = g(RR + h)²

h ≪ R: g(1 − 2hR)

g′ = g(1 − dR)

g′ = g − ω²R cos²λ

Field

I = GMr², I = −dVdr

sphere inside: GMrR³

shell inside: 0

x = √m₁ d√m₁ + √m₂

Potential & PE

V = −GMr, U = −GMmr

V_c = 32 V_s (sphere)

ΔU = mgh1 + hR

h = nR: nn + 1 mgR

Escape

v_e = √2GMR = √2gR

h = Rk²1 − k²

v′ = √v² − v_e²

v_e = √2 v_o

Satellites

v = √GMr, T = 2π√r³GM

L = m√GMr

K = GMm2r, E = −K, U = 2E

r_geo = (gR²ω²)^1/3

TOP MISTAKES

Avoid these = easy marks

MISTAKE ✗

Same g at height h and depth h

Using height h where r is needed

Gravity is zero in a satellite

Launch energy = orbital KE only

CORRECT ✓

d = 2h for the same g

r = R + h, always from the centre

N = 0, but gravity still acts

Energy = GMmR − GMm2r

Gravitation

Newton’s Law of Gravitation

8.1.1 Newton’s law of gravitation

8.1.2 Universal gravitational constant

8.1.3 Nature of the gravitational force

8.1.4 Force due to many particles

8.1.5 What distance to use

Corner

Acceleration Due to Gravity

8.2.1 Acceleration due to gravity

8.2.2 Which formula to use

8.2.3 What changes with g on a planet

8.2.4 Weighing the earth

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Variation of g

8.3.1 g above the surface

8.3.2 g below the surface

8.3.3 Comparing the two

8.3.4 Equator and poles

8.3.5 g at latitude λ

8.3.6 How fast must the earth spin?

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Gravitational Field

8.4.1 Gravitational field intensity

8.4.2 Where the field is zero

8.4.3 Link with V

8.4.4 Uniform sphere of mass M

8.4.5 Thin shell of mass M

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Gravitational Potential & Energy

8.5.1 Gravitational potential

8.5.2 Scalar addition

8.5.3 Thin spherical shell

8.5.4 Uniform sphere

8.5.5 Energy of position

8.5.6 Raising a body by height h

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Escape Velocity

8.6.1 Kinetic energy pays for the escape

8.6.2 Which formula to use

8.6.3 Maximum height reached

8.6.4 Speed far away

8.6.5 Speed on reaching the surface

8.6.6 Not from the surface

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Kepler’s Laws of Planetary Motion

8.7.1 Kepler’s laws

8.7.2 Areal velocity

8.7.3 Comparing planets

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Satellites: Orbital Velocity & Period

8.8.1 Speed in a circular orbit

8.8.2 h ≪ R

8.8.3 Horizontal velocity VH at a height

8.8.4 One revolution

8.8.5 Shortest period

8.8.6 If F ∝ 1 ÷ rⁿ

8.8.7 Particles moving under their own gravity

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Geostationary Satellite & Angular Momentum

8.9.1 From its period

8.9.2 Fixed over one place

8.9.3 Radius, height and speed

8.9.4 Another satellite of the same planet

8.9.5 Of a satellite

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Satellite Energy & Weightlessness

8.10.1 In a circular orbit

8.10.2 To free the satellite

8.10.3 From the surface into orbit

8.10.4 Two different jobs

8.10.5 Moving to a higher orbit

8.10.6 Weight is the reaction of the support

Corner

Just see this before the exam

Whole chapter on one page

Avoid these = easy marks

8.8.3 Horizontal velocity V_H at a height