SPI Physics · Gravitation

Gravitational Force, g and Satellites

Every two masses pull on each other with a force that grows with their masses and falls with the square of the distance between them; near the Earth that pull gives every body the same acceleration g.

All Gravitation notes

Why this matters

Five questions in the bank, three of them from the 2022 paper: the mass of the Sun from the Earth's pull, why g is larger at the poles, where g is least, the period of a geostationary satellite and the Earth's escape velocity. Three are EASY and two MODERATE. The escape velocity question turns on a unit (km/s or m/s), so learn the numbers with their units.

Concept 1 of 3: Newton's law of gravitation

The pull between two bodies gets stronger if either is heavier, and much weaker as they move apart: twice as far means a quarter of the pull. That is why the Sun, though huge, pulls you far less than the Earth under your feet.

Definition

Every two bodies attract each other. The force is proportional to the product of their masses and inversely proportional to the square of the distance between their centres.
G is the universal gravitational constant, $6.67 \times 10^{-11}\ \text{N m}^2/\text{kg}^2$ . It is the same everywhere in the universe.
Double the distance and the force becomes $\frac{1}{4}$ ; triple it and it becomes $\frac{1}{9}$ .
To find an unknown mass, rearrange: $M = \dfrac{F r^2}{G m}$ .

Universal law of gravitation

F = \frac{G m_1 m_2}{r^2} \qquad G = 6.67 \times 10^{-11}\ \text{N m}^2\,\text{kg}^{-2}

m₁, m₂the two masses, in kg
rdistance between their centres, in metres

Worked example

Two bodies of masses 100 kg and 60 kg are 2 m apart. Find the gravitational force between them.

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

SPI · 2022 · SPI Entrance Examination 2022 · Q91Moderate

Example 1 · Gravitation · Gravitational Force, g and Satellites

The mass of earth is

6 \times 10^{24}

kg. The distance between the earth and the sun is

1.5 \times 10^{11}

m. The gravitational force between the two is

3.5 \times 10^{22}

N then the mass of the sun is ____kg.

Halving the force when the distance doubles

The force goes with the square of the distance. Twice the distance gives a quarter of the force, not half.

Forgetting to square the distance

F = \frac{G m_1 m_2}{r^2}

the distance is squared. A distance of

3 \times 10^{5}

m becomes

9 \times 10^{10}

in the formula.

Concept 2 of 3: The value of g and how it changes over the Earth

g is the Earth's pull on each kilogram. Stand nearer the Earth's centre and the pull is stronger. The Earth is not a perfect ball: it is a little flattened at the poles, so the poles are nearer the centre than the equator is.

Definition

Near the Earth's surface every falling body speeds up at $g \approx 9.8\ \text{m/s}^2$ , whatever its mass.
From the law of gravitation, $g = \dfrac{GM}{R^2}$ , where M and R are the Earth's mass and radius. A smaller R gives a larger g.
The Earth bulges at the equator and is flattened at the poles. So g is greatest at the poles and least at the equator, and it decreases from pole to equator. The Earth's spin also lowers g most at the equator.
g also decreases as you go up (height) and as you go down a mine (depth). At the Earth's centre g is zero.
Weight $= mg$ changes from place to place; mass stays the same everywhere. On the Moon g is about one sixth of the Earth's.

Acceleration due to gravity

g = \frac{GM}{R^2} \approx 9.8\ \text{m/s}^2 \qquad W = mg

Worked example

A planet has the same mass as the Earth but half its radius. How does g on the planet compare with g on the Earth?

Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionEasy

Example 2 · Gravitation · Gravitational Force, g and Satellites

Gravity is more at the poles than in the equator because

Putting the largest g at the equator

The equator is farther from the Earth's centre than the poles, because of the bulge. So g is least at the equator and greatest at the poles.

Giving a reason that has nothing to do with gravity

Sunlight, ice cover and the Earth's magnetism do not change g. The reasons are the Earth's shape (distance from the centre) and its spin.

Saying mass changes on the Moon

Weight changes with g; mass, the amount of matter, does not. A 60 kg body is 60 kg on the Moon too.

Drill 1 more on the value of g and how it changes over the earth

Concept 3 of 3: Satellites and escape velocity

Throw a ball faster and it lands farther away. Throw it fast enough and the ground curves away as fast as the ball falls, so it keeps going round: that is a satellite. Throw it faster still and it never comes back: that is escape velocity.

Definition

A satellite is a body that goes round a planet. The Moon is the Earth's natural satellite and takes about 27 days to go round once.
A geostationary satellite goes round above the equator, from west to east, once every 24 hours, the same time the Earth takes to turn once. So it stays above the same place on the Earth. It is about 36,000 km up and is used for communication (TV, telephone).
Escape velocity is the least speed a body needs at the surface to leave the planet for good: $v_e = \sqrt{2gR}$ .
For the Earth, $v_e \approx 11.2$ km/s, which is 11,200 m/s. Watch the unit.

Escape velocity

v_e = \sqrt{2gR} \approx 11.2\ \text{km/s} = 11200\ \text{m/s} \text{ for the Earth}

Worked example

Find the Earth's escape velocity. Take

g = 9.8\ \text{m/s}^2

and

R = 6.4 \times 10^{6}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

SPI · 2022 · SPI Entrance Examination 2022 · Q85Easy

Example 3 · Gravitation · Gravitational Force, g and Satellites

The time period of a geo-stationary satellite is ____.

Reading 11.2 as m/s

The Earth's escape velocity is 11.2 km/s. In m/s it is 11,200. If a question asks in m/s, 11.2 is a thousand times too small.

Matching a geostationary satellite to a year

A geostationary satellite keeps pace with the Earth's spin, which takes a day, so its period is 24 hours. 365 days is the time the Earth takes to go round the Sun.

Drill 1 more on satellites and escape velocity

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Newton's law of gravitation
Universal law of gravitation
$F = \frac{G m_1 m_2}{r^2} \qquad G = 6.67 \times 10^{-11}\ \text{N m}^2\,\text{kg}^{-2}$
The value of g and how it changes over the Earth
Acceleration due to gravity
$g = \frac{GM}{R^2} \approx 9.8\ \text{m/s}^2 \qquad W = mg$
Satellites and escape velocity
Escape velocity
$v_e = \sqrt{2gR} \approx 11.2\ \text{km/s} = 11200\ \text{m/s} \text{ for the Earth}$

Watch out for (7)

Halving the force when the distance doubles
→ Newton's law of gravitation
Forgetting to square the distance
→ Newton's law of gravitation
Putting the largest g at the equator
→ The value of g and how it changes over the Earth
Giving a reason that has nothing to do with gravity
→ The value of g and how it changes over the Earth
Saying mass changes on the Moon
→ The value of g and how it changes over the Earth
Reading 11.2 as m/s
→ Satellites and escape velocity
Matching a geostationary satellite to a year
→ Satellites and escape velocity

Test yourself on a real paper

Sit a past SPI paper, timed and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Sit an SPI past paper Drill all 5 questions