Lesson 2 of 6Gravitational, Electric, and Magnetic Fields

Unit 3 · Lesson 2 of 6

Gravitational Potential Energy, Escape Speed and Kepler's Laws

Why gravitational potential energy far from Earth is written as a negative number, how much energy it takes to change orbits or leave a planet for good, and how Kepler's three laws describe every orbit in the solar system.

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Key things to remember

1Gravitational potential energy with zero at infinity: $E_{p} = - \frac{GM m}{r}$ ; it is negative because the object is bound.
2Change in potential energy between two radii: $Δ E_{p} = GM m (\frac{1}{r _{1}} - \frac{1}{r _{2}})$ , which reduces to $m g h$ near the surface.
3In a circular orbit, $E_{k} = \frac{GM m}{2 r}$ and $E_{t o t a l} = - \frac{GM m}{2 r}$ ; the binding energy is $\frac{GM m}{2 r}$ .
4Escape speed: $v_{esc} = \frac{2 GM}{r} = 2 v_{or bi t}$ , independent of the object's mass and launch direction.
5Kepler's laws: elliptical orbits with the Sun at one focus; equal areas in equal times; $T^{2} \propto r^{3}$ , so $\frac{T _{1}^{2}}{r _{1}^{3}} = \frac{T _{2}^{2}}{r _{2}^{3}}$ for the same central body.

In Grade 11 you used $E_{p} = m g h$ , and it works well for a ball tossed onto a roof. Try it on a rocket heading to the Moon, though, and it fails badly, because it assumes $g$ is the same at every height. Gravity fades with distance, so each extra kilometre of height costs a little less energy than the one before. Follow that idea all the way out and something surprising happens: the total energy needed to climb out of Earth's gravity completely is finite. That single fact explains escape speed, the energy cost of moving satellites between orbits, and why the energy of a planet in orbit is negative.

Gravitational potential energy far from a planet

Potential energy only means something as a difference, so we are free to choose where it is zero. Near the ground, the floor or the ground is convenient. For orbits there is no natural floor, so physicists set $E_{p} = 0$ when two masses are infinitely far apart, where they no longer affect each other at all. Bringing the masses closer lets gravity do positive work, so the potential energy drops below zero. Adding up the work done by the inverse-square force, which is the area under the force–distance graph, gives a remarkably simple result.

E_{p} = - \frac{GM m}{r} Δ E_{p} = GM m (\frac{1}{r _{1}} - \frac{1}{r _{2}})

r is measured from the planet's centre. Moving outward from r1 to r2 makes Ep less negative, so delta Ep is positive.

What the minus sign means

A negative potential energy says the object is bound: you would have to add energy to separate it from the planet. Picture a deep well with steep walls near the bottom that flatten out higher up. Earth's surface is near the bottom of Earth's well, at

E_{p} = - GM m / R_{E}

. Climbing out takes energy, and the closer you are to the planet, the deeper you sit. Near the surface,

Δ E_{p} = GM m (1/ r_{1} - 1/ r_{2})

reduces to almost exactly

m g h

, which is why the Grade 11 formula is a good local approximation.

Fig 1Field lines crowd together near Earth, where the field is strong and the energy well is steepest. They spread out with distance, so each kilometre farther out costs less energy.

Energy in a circular orbit

For a circular orbit, Universal Gravitation and Orbits showed that $v^{2} = GM / r$ . Multiply by $\frac{1}{2} m$ and the kinetic energy becomes $\frac{GM m}{2 r}$ , exactly half the size of the potential energy. Adding the two gives a total energy that is negative and equal to half the potential energy. The binding energy is the energy you would have to add to free the satellite completely, which is the size of that total. Notice what happens as $r$ grows: the kinetic energy falls (higher orbits are slower), but the potential energy rises twice as fast, so the total energy increases. To climb to a higher orbit, a satellite must fire its engines and add energy, yet it ends up moving more slowly.

E_{k} = \frac{GM m}{2 r} E_{t o t a l} = E_{k} + E_{p} = - \frac{GM m}{2 r} Δ E_{or bi t} = \frac{GM m}{2} (\frac{1}{r _{1}} - \frac{1}{r _{2}})

The last expression is the energy that must be added to move a satellite from a circular orbit of radius r1 to a larger one of radius r2.

Escape speed

Launch an object straight up and it slows as it climbs. If it starts with enough kinetic energy to make the total energy zero, it will just barely reach infinite distance, slowing toward a stop but never coming back. That launch speed is the escape speed. Setting $\frac{1}{2} m v^{2} - GM m / r = 0$ cancels the object's mass, so a pebble and a spacecraft need the same speed. The direction does not matter either, as long as the path does not hit the planet, because energy is a scalar. Comparing with orbital speed shows that escape speed is always $2$ times the circular-orbit speed at the same radius. Real rockets never actually reach $11.2 km/s$ at the launch pad: they burn their engines over several minutes and climb above the thick atmosphere first. Escape speed is the speed needed if all the energy is given at the start.

v_{esc} = \frac{2 GM}{r} = 2 v_{or bi t}

Escape speed from a distance r from the centre of a body of mass M. From the surface, r is the body's radius.

Body	Mass (kg)	Radius (m)	Escape speed from the surface
Moon	$7.35 \times 1 0^{22}$	$1.74 \times 1 0^{6}$	$2.37 km/s$
Mars	$6.42 \times 1 0^{23}$	$3.39 \times 1 0^{6}$	$5.03 km/s$
Earth	$5.97 \times 1 0^{24}$	$6.37 \times 1 0^{6}$	$11.2 km/s$
Jupiter	$1.90 \times 1 0^{27}$	$6.99 \times 1 0^{7}$	$60.2 km/s$
Sun	$1.99 \times 1 0^{30}$	$6.96 \times 1 0^{8}$	$618 km/s$

Computed with G = 6.67 × 10⁻¹¹ N·m²/kg². The Moon's low escape speed is one reason it has kept almost no atmosphere: fast-moving gas molecules leak away.

Kepler's three laws

Law of ellipses. Each planet moves in an ellipse with the Sun at one focus. A circle is the special case where the two foci coincide. Most planetary orbits are nearly circular; comets follow long, stretched ellipses.
Law of equal areas. A line from the Sun to the planet sweeps out equal areas in equal times. The planet therefore moves fastest at perihelion (closest approach) and slowest at aphelion (farthest point). In energy terms, near the Sun the potential energy is most negative, so the kinetic energy is largest.
Law of periods. The square of the period is proportional to the cube of the average orbital radius: $T^{2} \propto r^{3}$ . You already derived this for circular orbits as $T^{2} = \frac{4 π ^{2}}{GM} r^{3}$ , so the constant depends only on the central mass. For two planets of the same star, $\frac{T _{1}^{2}}{r _{1}^{3}} = \frac{T _{2}^{2}}{r _{2}^{3}}$ .

Weighing stars and defining black holes

Johannes Kepler found his laws from Tycho Brahe's careful observations of Mars, around 1609 to 1619, long before anyone knew why they worked; Newton later showed that all three follow from the inverse-square law. Because the constant in the third law is

4 π^{2} / (GM)

, measuring any orbit's radius and period gives the central mass. Earth's year and distance give the Sun's mass, and the same method weighs stars with exoplanets. Push escape speed to its limit and set

v_{esc} = c

: the radius

r = 2 GM / c^{2}

is the size a mass would need to be squeezed into for not even light to escape. For the Sun that is about

3 km

, the scale of a black hole.

Worked examples

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

Example 1

500 kg

satellite is lifted from Earth's surface to an altitude of

400 km

. Find the change in its gravitational potential energy using

E_{p} = - GM m / r

, and compare with the value

m g h

would give. Use

M_{E} = 5.97 \times 1 0^{24} kg

and

R_{E} = 6.37 \times 1 0^{6} m

Example 2

450 kg

satellite is in a circular orbit

400 km

above Earth's surface (

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

). Find its kinetic, potential and total energy, and its binding energy.

Example 3

Find the escape speed from Earth's surface, then from the

400 km

orbit of the previous example. How much extra speed must a spacecraft already in that orbit gain to escape?

Example 4

Earth orbits the Sun at an average radius of

1.50 \times 1 0^{11} m

with a period of

1.00

year. Mars orbits at an average radius of

2.28 \times 1 0^{11} m

. Use Kepler's third law to find the length of a Martian year.

Where marks get lost

Dropping the minus sign in $E_{p} = - GM m / r$ , which makes moving outward look like a loss of potential energy. Moving away from a planet always increases (makes less negative) the potential energy.
Using the altitude instead of the distance from the centre. Every $r$ in $- GM m / r$ , $2 GM / r$ and Kepler's third law is measured from the centre of the central body.
Thinking escape speed depends on the mass of the object or the direction it is launched. Mass cancels and energy is a scalar, so neither matters (as long as the path misses the planet).
Assuming a satellite in a higher orbit moves faster because it has more energy. Its total energy is higher, but its kinetic energy and speed are lower.
Applying $T^{2} / r^{3}$ ratios between objects orbiting different central bodies, such as the Moon and Mars. The constant $4 π^{2} / (GM)$ depends on the mass being orbited.

Flashcards 9

Gravitational potential energy (general)

Why is gravitational Ep negative?

Kinetic energy in a circular orbit

Total energy of a circular orbit

+ 5 more

Practice

What is the escape speed from the surface of the Moon (

M = 7.35 \times 1 0^{22} kg

R = 1.74 \times 1 0^{6} m

), in km/s? Give three significant figures.

km/s

What is the escape speed from a spacecraft's position

400 km

above Earth's surface, in km/s? Use

M_{E} = 5.97 \times 1 0^{24} kg

and

R_{E} = 6.37 \times 1 0^{6} m

, and give three significant figures.

km/s

1200 kg

satellite orbits Earth in a circle of radius

7.00 \times 1 0^{6} m

. What is its binding energy, in gigajoules (GJ)? Use

M_{E} = 5.97 \times 1 0^{24} kg

and give three significant figures.

How much energy must be added to move a

1000 kg

satellite from a circular orbit of radius

7.0 \times 1 0^{6} m

to one of radius

1.4 \times 1 0^{7} m

, in gigajoules? Use

M_{E} = 5.97 \times 1 0^{24} kg

and give three significant figures.

Jupiter orbits the Sun at an average distance 5.20 times Earth's. Using Kepler's third law, what is Jupiter's period, in Earth years? Give three significant figures.

years

Why is gravitational potential energy written as a negative number in

E_{p} = - GM m / r

A satellite is boosted from a low circular orbit to a higher circular orbit. Compared with before, which statement is true?

Two probes, one of

100 kg

and one of

2000 kg

, are launched from Earth's surface. How do their escape speeds compare?

At what point in its elliptical orbit does a planet move fastest? Give the one-word name.

Order these bodies by escape speed from their surface, from lowest to highest.

1The Sun
2Jupiter
3Earth
4The Moon
5Mars

Match each of Kepler's laws or results to what it states.

First law

Second law

Third law

Constant in the third law

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