Lesson 1 of 6Gravitational, Electric, and Magnetic Fields

Unit 3 · Lesson 1 of 6

Universal Gravitation and Orbits

The law that pulls an apple down and holds the Moon in orbit, what gravitational field strength really measures, and how to work out the speed and period of any satellite.

4 min read10 practice questionsAsk AI about this →

Key things to remember

1Universal gravitation: $F_{g} = \frac{G m _{1} m _{2}}{r ^{2}}$ , with $r$ measured centre to centre and $G = 6.67 \times 1 0^{- 11} N \cdot m^{2} / kg^{2}$ .
2Gravitational field strength: $g = \frac{F _{g}}{m} = \frac{GM}{r ^{2}}$ , in $N/kg$ .
3For a circular orbit gravity is the centripetal force, so $v = \frac{GM}{r}$ and $T = \frac{2 π r}{v}$ .
4Orbital radius is Earth's radius plus altitude: $r = R_{E} + h$ . The satellite's own mass does not affect its speed or period.
5A geostationary satellite orbits above the equator with a period of one day, at $r \approx 4.22 \times 1 0^{7} m$ .
6Apparent weightlessness happens when an object and its surroundings share the same free-fall acceleration, so no support force acts.

In Grade 11 you treated gravity as a constant $9.81 N/kg$ pulling straight down. That works on a football field, but it cannot explain why a satellite circles the Earth without falling in, or why astronauts float. The bigger picture is that every mass attracts every other mass, and the strength of that pull fades with distance. Once you have that one law, orbits, satellite periods and apparent weightlessness all follow from the circular-motion ideas you already know.

Newton's law of universal gravitation

F_{g} = \frac{G m _{1} m _{2}}{r ^{2}} G = 6.67 \times 1 0^{- 11} N \cdot m^{2} / kg^{2}

r is the distance between the centres of the two masses, not the gap between their surfaces.

Always attractive. There is no negative mass, so gravity never pushes objects apart.
Mutual. The Earth pulls you with exactly the same size of force that you pull the Earth, as Newton's third law demands. The Earth barely accelerates because its mass is enormous.
Inverse square. Double the centre-to-centre distance and the force drops to one quarter; triple it and it drops to one ninth.
Tiny for everyday masses. $G$ is so small that two $70 kg$ people standing $1.5 m$ apart attract each other with about $1.5 \times 1 0^{- 7} N$ . Gravity only becomes large when at least one mass is planet-sized.

Gravitational field strength

A field is a way of describing what a mass does to the space around it before anything else arrives. The gravitational field strength $g$ at a point is the gravitational force per kilogram that a small test mass would feel there: $g = F_{g} / m$ . Dividing Newton's law by the test mass leaves $g = GM / r^{2}$ , where $M$ is the mass creating the field. Put in Earth's mass $5.97 \times 1 0^{24} kg$ and radius $6.37 \times 1 0^{6} m$ and you get $9.81 N/kg$ , the familiar surface value. Move farther out and $g$ shrinks with the square of the distance from Earth's centre.

Fig 1Field lines point toward Earth's centre and spread apart with distance, so the field weakens. In orbit, gravity is the centripetal force.

Near-Earth fields and distant fields

Far from the Earth the field lines are radial: they point at the centre and spread out, so the field weakens as

1/ r^{2}

. Close to the surface, over a region a few kilometres across, the lines are almost parallel and evenly spaced, so the field is effectively uniform at

9.81 N/kg

straight down. That is why the Grade 11 constant-

g

model works for projectiles but fails for satellites.

Satellites: gravity as the centripetal force

A satellite in a circular orbit is always falling toward the Earth, but it is also moving sideways fast enough that the ground curves away beneath it at the same rate. The only force acting is gravity, so gravity alone supplies the centripetal force. Setting the two equal gives the orbital speed, and dividing the circumference by that speed gives the period.

\frac{GM m}{r ^{2}} = \frac{m v ^{2}}{r} \Rightarrow v = \frac{GM}{r} T = \frac{2 π r}{v} \Rightarrow T^{2} = \frac{4 π ^{2} r ^{3}}{GM}

M is the mass being orbited, r = Earth radius plus altitude. The satellite's own mass cancels.

Altitude above the surface	Orbital radius r	Orbital speed	Period
$400 km$ (low Earth orbit)	$6.77 \times 1 0^{6} m$	$7.67 km/s$	$92.4 min$
$800 km$	$7.17 \times 1 0^{6} m$	$7.45 km/s$	$101 min$
about $35900 km$ (geostationary)	$4.22 \times 1 0^{7} m$	$3.07 km/s$	$24.0 h$

Computed from the orbit equations with this course's values of G and Earth's mass. Higher orbits are slower and take longer.

Geostationary orbits

A satellite orbiting above the equator, in the direction the Earth spins, with a period of one day stays over the same spot on the ground. Dish antennas can then point at it permanently, which is why communication and weather satellites use this orbit. Because every geostationary satellite must have the same period, they all share the same radius, about

4.22 \times 1 0^{7} m

from Earth's centre. (Strictly, the Earth turns once relative to the stars in about 23 h 56 min, which shifts the radius only slightly.)

Apparent weightlessness is not zero gravity

400 km

altitude

g

is still about

8.69 N/kg

, roughly 89% of the surface value. Astronauts float because they and their spacecraft are falling around the Earth together with the same acceleration, so the floor never has to push on them. A scale reads the normal force, and the normal force is zero. That is apparent weightlessness: gravity is fully present, the support force is not.

Beyond Newton

Newton's law predicts orbits extremely well, but it describes gravity as a force without explaining how mass produces it. Einstein's general theory of relativity describes gravity instead as the curvature of spacetime by mass and energy, and it is needed near very massive objects such as black holes. Orbit calculations also underpin the evidence for dark matter: stars in the outer parts of galaxies orbit faster than the visible matter alone can explain.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

450 kg

satellite orbits

400 km

above Earth's surface. Find the gravitational field strength at that altitude and the gravitational force on the satellite.

Example 2

Find the orbital speed and the period, in minutes, of the satellite in the previous example, orbiting at

r = 6.77 \times 1 0^{6} m

Example 3

Find the orbital radius and the altitude of a geostationary satellite, taking its period as

24.0 h

Example 4

A moon orbits a distant planet in a circle of radius

2.0 \times 1 0^{8} m

with a period of

3.0

days. Find the mass of the planet.

Where marks get lost

Using the altitude as $r$ . The distance in $F_{g} = G m_{1} m_{2} / r^{2}$ and $v = GM / r$ is measured from Earth's centre, so add Earth's radius first.
Forgetting to square $r$ , or halving the force when the distance doubles. It is an inverse square: doubling $r$ quarters the force.
Putting the satellite's mass into $v = GM / r$ . $M$ is the mass being orbited; the satellite's mass cancels.
Saying there is no gravity in orbit. Gravity in low Earth orbit is nearly 90% of its surface value; it is the support force that is missing.
Leaving the period in hours or days when using $G$ in SI units. Convert to seconds, then convert back at the end.

Flashcards 8

Law of universal gravitation

Value of G

Gravitational field strength

Orbital speed

+ 4 more

Practice

The distance between the centres of two asteroids triples. What happens to the gravitational force between them?

Two cars, of mass

1000 kg

and

2000 kg

, are parked with their centres

2.0 m

apart. What is the gravitational force between them, in micronewtons (μN)? Give three significant figures.

μN

A probe is at an altitude equal to two Earth radii above the surface. What is Earth's gravitational field strength there, in N/kg? Give two decimal places.

N/kg

What is the orbital speed of a satellite

800 km

above Earth's surface, in km/s? Give three significant figures.

km/s

What is the period, in minutes, of the satellite orbiting

800 km

above Earth's surface? Give three significant figures.

min

A second satellite with twice the mass is placed in the same circular orbit as the first. How does its orbital speed compare?

Which statement best explains why astronauts float inside a spacecraft in low Earth orbit?

What name is given to an orbit above the equator with a period of one day, so that the satellite stays above the same point on the ground?

Match each quantity to its expression.

A. $GM / r$
B. $2 π r / v$
C. $G m_{1} m_{2} / r^{2}$
D. $GM / r^{2}$

Gravitational force between two masses

Gravitational field strength

Orbital speed

Orbital period

Order these Earth orbits from shortest period to longest period.

1The Moon's orbit, about $384000 km$ from Earth
2A geostationary orbit, about $35900 km$ above the surface
3A circular orbit $20000 km$ above the surface
4A circular orbit $400 km$ above the surface
5A circular orbit $2000 km$ above the surface

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