Physics notes, chapter 3

Gravitation

Textbook Physics Form 4 KSSM (DLP), pages 76–116

What this chapter covers

State Newton's Universal Law of Gravitation and use F = Gm₁m₂/r² (3.1)
Relate g to G and calculate g on the Earth, at a height h and on other bodies (3.1)
Explain centripetal force and use F = mv²/r and a = v²/r for satellites and planets (3.1)
Calculate the mass of the Earth or the Sun from an orbit using M = 4π²r³/GT² (3.1)
State Kepler's three laws and use T₁²/T₂² = r₁³/r₂³ (3.2)
Find the linear speed of a satellite, v = √(GM/r), and compare geostationary and non-geostationary satellites (3.3)
Define escape velocity, calculate it with v = √(2GM/R) and explain its benefits and implications (3.3)

3.1Newton's universal law of gravitation

Gravitational force is a universal force: it acts between any two bodies in the universe, for example the Sun and the Earth, the Earth and the Moon, or a durian and the Earth.

In 1687 Isaac Newton showed that the gravitational force, F, between two bodies depends on two things: it is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.

Newton's Universal Law of Gravitation: F = Gm₁m₂ / r², where r is the distance between the centres of the two bodies and G = 6.67 × 10⁻¹¹ N m² kg⁻² (the gravitational constant).

Larger masses give a larger force.
Doubling the distance makes the force one quarter.
Two people standing 1 m apart feel almost no pull on each other (50 kg and 70 kg gives only about 2.3 × 10⁻⁷ N) because their masses are tiny.

A falling leaf and the Earth pull on each other with the same force. We only see the leaf move because the Earth's mass is enormously larger, so its acceleration is negligible.

Worked example: force on a durian

A durian of mass 2.0 kg is at the Earth's surface. Mass of the Earth = 5.97 × 10²⁴ kg, r = 6.37 × 10⁶ m.
F = (6.67 × 10⁻¹¹)(5.97 × 10²⁴)(2.0) ÷ (6.37 × 10⁶)² = 19.6 N (this is its weight, mg with g = 9.81).

Two bodies attract each other with force F. What is the force if both masses are doubled and the distance is also doubled?

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Masses: × 2 × 2 = × 4. Distance: ÷ 2² = ÷ 4. Net effect × 4 ÷ 4 = F (unchanged).

Gravitational acceleration, g

Weight is a gravitational force, so mg = GMm/r². The mass m of the object cancels:

g = GM / r². At the surface r = R; at height h above the surface r = R + h.

g depends on the mass of the planet and the distance from its centre, not on the mass of the object.
Earth's surface: 6.67 × 10⁻¹¹ × 5.97 × 10²⁴ ÷ (6.37 × 10⁶)² = 9.81 m s⁻².
A satellite at h = 480 km: g = GM/(R + h)² = 8.49 m s⁻², a little less than at the surface.

Graph of g against r: rises in a straight line to R, then falls as 1 over r squared — Variation of g with distance r from the centre of the Earth. Inside the Earth g is directly proportional to r; outside it is inversely proportional to r². At R, g = 9.81; at 2R, 2.45; at 3R, 1.09 m s⁻². (After textbook Figure 3.11, p. 84)

r is measured from the centre of the planet, not from its surface. For a satellite use r = R + h.

Why do astronauts need to know g for another planet before landing?

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It tells the weight (gravitational force) on objects and people there, so they can design landing gear, rocket thrust and equipment, and judge the effects of low or high gravity on the body (e.g. bone density, blood pressure).

Centripetal force and satellites

Centripetal force is the force on a body in circular motion that acts towards the centre of the circle. F = mv²/r, and centripetal acceleration a = v²/r.

Even at constant speed the direction keeps changing, so the body accelerates. In the textbook activity the tension in the string is the centripetal force: more weights on the string means a faster whirl, and a smaller radius needs a larger force. For a satellite or planet, gravitational force acts as the centripetal force.

Worked example: hammer throw

Iron ball m = 7.2 kg, v = 20 m s⁻¹, r = 1.8 m.
F = mv²/r = 7.2 × 20² ÷ 1.8 = 1 600 N.

Mass of the Earth or the Sun

Equate gravitational force and centripetal force for an orbiting body, then use v = 2πr/T:

M = 4π²r³ / GT². You need the radius and period of an orbit around the body. The mass of the orbiting body cancels.

Worked example: mass of the Earth from the Moon

r = 3.83 × 10⁸ m, T = 2.36 × 10⁶ s.
M = 4π²(3.83 × 10⁸)³ ÷ [(6.67 × 10⁻¹¹)(2.36 × 10⁶)²] = 5.97 × 10²⁴ kg.

Put T in seconds (1 day = 86 400 s) when using G. Cube r, then divide; do not forget the square on T.

3.2Kepler's Laws

Law	Statement	What it means
First (law of orbits)	Planets move in elliptical orbits with the Sun at one focus	The distance to the Sun changes; most planetary orbits are almost circular
Second (law of areas)	A line joining a planet to the Sun sweeps out equal areas in equal times	A planet moves faster when it is nearer the Sun
Third (law of periods)	T² is directly proportional to r³	Planets further from the Sun take longer to orbit (Earth 1 year, Saturn 29.5 years)

From gravitational force = centripetal force: T² = (4π²/GM)r³. The same law applies to satellites around the Earth, with M the Earth's mass. To compare two orbits around the same body:

T₁² / T₂² = r₁³ / r₂³ (use the same unit for both periods).

Worked example: orbit of Mars

Earth: r₁ = 1.50 × 10¹¹ m, T₁ = 1.00 year. Mars: T₂ = 1.88 years.
r₂³ = r₁³ × (T₂/T₁)², so r₂ = 1.50 × 10¹¹ × 1.88^2/3 = 2.28 × 10¹¹ m.

Saturn's orbit radius is 1.43 × 10¹² m. Earth's is 1.50 × 10¹¹ m. Find Saturn's period in years.

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T = (1.43 × 10¹² ÷ 1.50 × 10¹¹)^3/2 = 9.53^1.5 = 29.4 years.

3.3Man-made satellites

For a circular orbit of radius r = R + h: v = √(GM/r). The speed depends on the planet's mass and the orbit radius only, not on the satellite's mass.

Higher orbits mean slower satellites. GPS satellite at 20 200 km: r = 2.657 × 10⁷ m, v = 3.87 × 10³ m s⁻¹. If a satellite's speed drops below the required value, it falls to a lower orbit and may burn up in the atmosphere.

Earth with geostationary orbit, satellite A above point x and satellite B in a different orbit — Satellite A is geostationary (above the same place on Earth); satellite B is not.

Geostationary and non-geostationary satellites: textbook Figure 3.36, p. 105.

	Geostationary	Non-geostationary
Orbital period	24 hours (same as Earth's rotation)	Shorter or longer than 24 hours
Direction	Same as Earth's rotation	Need not be the same
Position	Always above the same place	Above different places at different times
Examples and uses	MEASAT, communication	GPS, ISS, weather and imaging (e.g. RazakSAT)

Escape velocity

Escape velocity is the minimum velocity an object needs at the surface to overcome the gravitational force and escape into outer space. v = √(2GM/R).

It depends on the mass and radius of the planet, not on the mass of the object. Escape velocity from the Earth is 11.2 km s⁻¹ (11 200 m s⁻¹).
Benefit: air molecules (about 500 m s⁻¹) cannot escape, so the Earth keeps its atmosphere; aircraft fly high without leaving the Earth.
Implication: rockets need huge amounts of fuel to reach this speed.
Mars has a low escape velocity (about 5.0 km s⁻¹) and thin air; Jupiter's is about 60 km s⁻¹.

Worked example: Moon and Sun

Moon: g = 1.62 m s⁻², v = 2.37 × 10³ m s⁻¹. Sun: g = 274 m s⁻², v = 6.18 × 10⁵ m s⁻¹. The Sun is far more massive, so both values are much higher.

Do not confuse the two square roots: orbital speed v = √(GM/r), escape velocity v = √(2GM/R). Escape velocity is √2 times the surface orbital speed.

Spacecraft X has mass 1 500 kg and spacecraft Y has mass 2 000 kg. Do they need different escape velocities from the Earth?

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No. Escape velocity is independent of the mass of the object; it depends only on the mass and radius of the Earth.