1 2 3 4 5 6

0:006 lessons across the whole video50:03

You will be able to

Derive the orbital speed and period of a body in circular orbit from Newton's law of gravitation
Apply conservation of energy to relate kinetic and gravitational potential energy in an orbit
Calculate the escape velocity from a planet's surface using GM and radius
Explain how Kepler's laws describe planetary motion and connect them to Newtonian gravity
Classify orbits as bound or unbound from the total mechanical energy
Distinguish elliptical, circular, and escape trajectories using energy and angular momentum

Good to know first

Newton's second law and centripetal acceleration
Newton's law of universal gravitation
Conservation of mechanical energy and gravitational potential energy
Basic calculus for integrating or manipulating energy expressions

Syllabus

6 lessons and a final project

1
Kepler's laws and the orbit problem: a small planet around a huge sun Free
0:00–9:009 min of video
This lesson sets up the orbit problem: a small mass m moving around a much heavier mass M, held by gravity. It shows how gravity supplies the centripetal force that keeps a planet on its path, and introduces the energy idea behind escape velocity.
4 formulas3 exercises4 questions
2
Circular orbits: gravity supplies the centripetal force Beyond Pass
9:00–20:4012 min of video
A satellite on a circular orbit is held in its path by gravity, which plays the role of the centripetal force. Setting gravity equal to mv²/r gives the orbital speed, and from it the period.
3 formulas2 exercises3 questions
3
Orbital speed and period via Kepler's third law Beyond Pass
20:40–27:207 min of video
This lesson shows how Kepler's third law links the period of a circular orbit to its radius, and how to combine it with gravity to get the orbital speed. You will practise these two formulas on a satellite example.
1 code3 formulas3 build steps3 exercises3 questions
4
Total energy of an orbit with gravitational potential Beyond Pass
27:20–35:008 min of video
This lesson shows how the total energy of an orbiting object is kinetic plus gravitational potential energy, and why that sum decides whether the orbit is bound or escapes. It also gives a feel for the sizes of energy and power involved, using the lecture's examples.
3 exercises3 questions
5
Deriving escape velocity from energy conservation Beyond Pass
35:00–42:007 min of video
An object launched from Earth's surface can escape gravity forever if it has enough kinetic energy to bring its total energy to zero at infinity. Energy conservation turns this into a single formula for the escape speed.
2 formulas2 exercises3 questions
6
Elliptical Orbits, Bound and Unbound Motion Beyond Pass
42:00–50:038 min of video
This lesson shows how an orbit's energy decides its shape: bound orbits are closed ellipses, while unbound orbits escape to infinity. It also summarises the main ideas of orbital motion from the lecture.
2 exercises3 questions

Final project

Mission Planner: From Circular Orbit to Escape

Build a small calculator and short report that plans a satellite mission around a planet using only $G M$ and radius. It should connect circular orbit speed, period, energy, and escape speed, and explain why each result follows from Newtonian gravity and Kepler's laws.

Its 7 tasks open with Beyond Pass.

Free preview · Lesson 1 in full

Kepler's laws and the orbit problem: a small planet around a huge sun

This lesson sets up the orbit problem: a small mass m moving around a much heavier mass M, held by gravity. It shows how gravity supplies the centripetal force that keeps a planet on its path, and introduces the energy idea behind escape velocity.

0:00 to 9:00 in the video

Scratchpad

Draw and write next to the video.

The idea

Lewin begins with the question of how a planet stays in orbit. Picture a small mass m (a planet) going around a much heavier mass M (the Sun, written M with a Sun symbol on the board). Because m is much, much smaller than M, we can treat M as sitting still and let the small mass do all the moving. The board shows the sketch: a roughly circular path of radius R around the heavy body, with the small body at one point.

The key idea is that gravity is the only force that matters here, and it points from m straight toward M. Gravity is the force that bends the path. Without it, the planet would fly off in a straight line, as Newton's first law says. On the board he writes F_grav = F_cent: the gravitational pull is exactly the centripetal force that keeps the planet turning in a circle of radius R.

Setting those two forces equal gives the speed of a circular orbit. Gravity pulls with strength G M m / R squared, and the centripetal force needed for speed v is m v squared over R. Cancelling one m and one R gives v squared = G M / R, so the speed depends on the heavy mass and the radius, not on the planet's own mass.

The same board also starts an energy approach. The total energy E is kinetic energy plus gravitational potential energy: E = (1/2) m v^2 - G M m / r. Lewin writes this for two positions, one at the start and one at the end of the motion, and sets them equal, because energy is conserved. This energy bookkeeping leads to two cases: if the total energy is negative, the body is bound and follows an orbit; if it is zero or positive, the body can escape.

The escape velocity is the speed at which the total energy is exactly zero when the body starts at the surface of the heavy body. Lewin works it out at the Earth's surface, obtaining v_esc = sqrt(2 G M / R), which is about 11.2 km/s for the Earth. He then notes that the escape speed for a circular orbit is smaller by a factor of the square root of two, so the two speeds are related but different.

Key points

Treat the heavy mass M as fixed when m is much smaller; only the small body moves. 4:51
Gravity is the centripetal force: G M m / R^2 = m v^2 / R. 5:00
For a circular orbit, v^2 = G M / R, so the planet's own mass does not matter. 6:11
Total energy E = (1/2) m v^2 - G M m / r is conserved along the path. 2:00
Escape velocity is v = sqrt(2 G M / R); it is about 11 km/s from the Earth's surface. 8:00

Words to know

Orbit: The curved path a small body follows around a much heavier body because gravity keeps pulling it toward that body.
Centripetal force: The inward force that keeps a body moving in a circle; here it is supplied by gravity.
Gravitational potential energy: The energy stored in the pull between two masses, written as -G M m / r for distance r.
Bound orbit: A path where the total energy is negative, so the body cannot get infinitely far away.
Escape velocity: The minimum starting speed from a surface needed for the total energy to reach zero, letting the body leave forever.

Formulas

F_{g r a v} = F_{c e n t} \Rightarrow \frac{G M m}{R^{2}} = \frac{m v^{2}}{R}

Gravity pulls with exactly the force needed to keep the small mass moving in a circle.

$G$: the universal gravitational constant
$M$: the mass of the heavy body (Sun or Earth)
$m$: the mass of the small orbiting body
$R$: the radius of the circular orbit
$v$: the speed of the small body

v^{2} = \frac{G M}{R}

For a circular orbit, the speed squared equals G times M divided by the radius.

$v$: orbital speed
$R$: orbit radius

E = \frac{1}{2} m v^{2} - \frac{G M m}{r}

Total energy is kinetic energy plus gravitational potential energy, and it stays constant along the path.

$E$: total mechanical energy
$r$: distance between the two bodies

v_{e s c} = \sqrt{\frac{2 G M}{R}}

The escape speed from a surface of radius R is the square root of two times G M over R.

$v_{e s c}$: escape velocity
$R$: radius of the heavy body (for example, the Earth)

Practice

Calculate Solution checked

Take G M = 4.0 × 10^14 m^3/s^2 for a heavy body and an orbit radius R = 2.0 × 10^7 m. Find the circular orbital speed v using v^2 = G M / R.

Hint

Compute GM/R first, then take the square root.

Solution

Compute v^2 = GM / R = (4.0 × 10^14) / (2.0 × 10^7).
v^2 = 2.0 × 10^7 m^2/s^2.
v = sqrt(2.0 × 10^7) ≈ 4.47 × 10^3 m/s.

v ≈ 4.5 × 10^3 m/s (about 4.5 km/s)

Calculate Solution checked

Using the same GM = 4.0 × 10^14 m^3/s^2 and R = 2.0 × 10^7 m, find the escape velocity v_esc = sqrt(2GM/R).

Hint

The escape speed is the circular speed times the square root of 2.

Solution

2GM/R = 2 × 2.0 × 10^7 = 4.0 × 10^7 m^2/s^2.
v_esc = sqrt(4.0 × 10^7) ≈ 6.32 × 10^3 m/s.

v_esc ≈ 6.3 × 10^3 m/s

Explain Solution checked

In your own words, explain why the mass of the small planet does not appear in the formula for the circular orbital speed.

Hint

Look at the two forces: gravity has m in it, and the centripetal force has m in it.

Solution

Gravity on m is G M m / R^2 and the centripetal force needed is m v^2 / R.
Both sides contain m, so it cancels, leaving v^2 = G M / R.
Any small body at the same radius moves at the same speed, whatever its mass.

Both the gravitational force and the force needed to turn in a circle are proportional to m, so m cancels; the speed depends only on M and R.

On your scratchpad

Draw a large circle with a heavy dot M in the centre and a small dot m on the circle. Draw an arrow from m to M labelled F_grav, and a tangent arrow labelled v. Write beside it: F_grav = F_cent, so G M / R^2 = v^2 / R.

Quiz

In the orbit sketch, what force keeps the small mass m moving in a circle around M?
1. Friction with space
2. Gravity pulling m toward M
3. The rocket thrust of m
4. The magnetic force of M
Show the answer
B. Gravity pulling m toward M · Gravity points from m toward M, and it supplies the centripetal force.
5:00 watch the moment
For a circular orbit, what does the orbital speed depend on?
1. Only the mass m of the planet
2. The heavy mass M and the radius R
3. The color of the planet
4. Only the time of day
Show the answer
B. The heavy mass M and the radius R · From G M m / R^2 = m v^2 / R, v^2 = G M / R, which depends on M and R only.
6:11 watch the moment
Why can we treat the heavy mass M as not moving when m is much smaller?
1. Because M has no gravity
2. Because M is so large that its motion is negligible compared with m's
3. Because m is always at rest
4. Because energy is not conserved
Show the answer
B. Because M is so large that its motion is negligible compared with m's · Since m is much smaller than M, the heavy body barely moves, so the small body does almost all the moving.
4:51 watch the moment
What is the escape velocity from the Earth's surface, roughly?
1. 11 km/s
2. 11 m/s
3. 110 km/s
4. 1.1 km/s
Show the answer
A. 11 km/s · Lewin's board gives about 11.2 km/s, as v = sqrt(2GM/R) for the Earth.
8:00 watch the moment

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Lessons 2–6 and the final project

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Made by Beyond from the video “Lec 14: Orbits and Escape Velocity | 8.01 Classical Mechanics, Fall 1999 (Walter Lewin)” by For the Allure of Physics. The video plays in YouTube's own player. Watch it on YouTube · All lessons

6 lessons and a final project

Kepler's laws and the orbit problem: a small planet around a huge sun Free

Circular orbits: gravity supplies the centripetal force Beyond Pass

Orbital speed and period via Kepler's third law Beyond Pass

Total energy of an orbit with gravitational potential Beyond Pass

Deriving escape velocity from energy conservation Beyond Pass

Elliptical Orbits, Bound and Unbound Motion Beyond Pass

Mission Planner: From Circular Orbit to Escape

Kepler's laws and the orbit problem: a small planet around a huge sun

The idea

Key points

Words to know

Formulas

Practice

On your scratchpad

Quiz

Lessons 2–6 and the final project