Press Next (or Play) to walk through each story one small step at a time. The numbers come last.
1. The throw that never lands
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On a clear evening you can see the International Space Station cross the sky as a bright, steady dot. It is only
about 420 km up, closer than many cities are to each other. Gravity there is still about 90 % as strong as on the ground.
So why does it not fall down? In fact it is falling, all the time. It just moves sideways so fast that the
ground curves away beneath it as fast as it falls.
How fast must it go to keep circling, how long does one lap take, and what happens to its energy if it goes faster or slower?
2. Fast when close, slow when far
A satellite on a stretched (elliptical) orbit moves from its closest point to its farthest point. What happens to its speed?
Another example: The Moon is falling too
3. Speed up: higher orbit, or escape
4. Heavy or light, same ride
Two satellites circle Earth at the same height. One has 10 times the mass of the other. How do their speeds compare?
Another example: Stuck in orbit: the energy
5. Higher is slower
6. Floating while falling
Why do astronauts float inside the space station?
Another example: Ben's GPS satellite
Ben reads that a GPS satellite circles Earth at r = 2.66 × 10⁷ m from Earth's centre. How fast does it move, and how long is one lap? (GM = 3.98 × 10¹⁴ m³/s²)
Gravity is the centre-seeking force: GMm/r² = mv²/r. The satellite's mass m cancels, so v = √(GM/r).
v = √(3.98 × 10¹⁴ ÷ 2.66 × 10⁷) = √(1.50 × 10⁷) ≈ 3870 m/s, about 3.9 km/s.
One lap is a circle of length 2πr = 2π × 2.66 × 10⁷ ≈ 1.67 × 10⁸ m.
T = 2πr ÷ v = 1.67 × 10⁸ ÷ 3870 ≈ 4.32 × 10⁴ s, which is about 12 hours. Two laps a day.
Check: it is higher than the space station (6.79 × 10⁶ m), so it should be slower (3.9 km/s < 7.66 km/s) with a longer lap (12 h > 93 min). It is.
7. Check yourself
Think of your answer first, then tap to see it.
a) Astronauts float inside the space station. Is there gravity up there?
Show answer
Yes, about 90 % as strong as on the ground. The astronauts and the station are all falling together around Earth, so nothing presses them to the floor.
b) A heavy satellite and a light one circle Earth at the same distance r. Which one must move faster?
Show answer
Neither: they move at the same speed. In GMm/r² = mv²/r the satellite's mass m cancels, so v = √(GM/r).
c) At its closest point a satellite is 1.0 × 10⁷ m from Earth's centre and moves at 7.3 km/s. At its farthest point it is 2.0 × 10⁷ m away. How fast is it going there?
Show answer
m × v × r stays the same, so v = 7.3 × (1.0 ÷ 2.0) ≈ 3.7 km/s. Twice as far, half as fast.
Already know this?
Three quick questions. Get all three right on the first try and you can skip ahead.
Satellite B circles Earth at 4 times the orbit radius of satellite A. How does B's speed compare with A's?
A satellite moves on an elliptical orbit. Why does its angular momentum about Earth's centre stay the same?
A satellite circles Earth. Take U = 0 very far away. What is the sign of its total energy K + U?
The idea
Picture the space station circling Earth. It is falling toward Earth all the time. But it also moves sideways
very fast. So it keeps missing the ground, and goes round and round.
Some orbits are circles. Others are stretched ovals, called ellipses. On an ellipse the satellite speeds up as it
swings close to Earth. It slows down as it climbs far away.
The system. We study the satellite and the planet together as one system. Gravity between them is an
internal force. Gravity always points toward the planet's center, so it makes no torque about that center.
That gives us two conservation laws for every orbit:
Angular momentum about the planet's center is constant. For a satellite, L = m v r⊥. When the satellite is
closer, it must move faster. At the closest and farthest points v is at right angles to r, so r1v1 = r2v2.
Mechanical energy of the satellite-planet system is constant. E = K + U stays the same. Kinetic energy turns into
gravitational potential energy as the satellite climbs, and back again as it comes closer.
Gravitational potential energy far from the ground. Near the ground we used U = mgh. In space we must use
U = −G M m / r
Trap: using U = mgh for a satellite in orbit. Instead: g is not 9.8 m/s² up there and changes with r. Use U = −GMm/r, with r measured from the planet's center.
where r is the distance between the centers. U is zero when the objects are infinitely far apart and negative
everywhere else. Moving farther away makes U less negative, so it increases.
Trap: using the height above the ground (like 420 km) as r. Instead: r is center to center: r = RE + height = 6.37 × 10⁶ m + 0.42 × 10⁶ m = 6.79 × 10⁶ m.
Circular orbits. In a circle, gravity is the centripetal force: GMm/r² = mv²/r. Solve for v:
v = √(GM / r) T = 2πr / v = 2π √(r³ / GM)
Trap: "astronauts float because there is no gravity up there". Instead: at 420 km gravity is still about 90 % of its ground value. They float because they and the station fall together around Earth (free fall).
The satellite's own mass m cancels. A 1 kg probe and a 400 000 kg station at the same r have the same speed.
Higher orbits are slower and take longer: T² is proportional to r³ (Kepler's third law).
Energy in a circular orbit: K = GMm/(2r), U = −GMm/r, so E = K + U = −GMm/(2r). Notice E = −K and U = 2E.
Trap: "a higher orbit is slower, so it has less energy". Instead: higher is slower (less K), but U rises more, so total E = −GMm/(2r) is less negative: a higher orbit has MORE total energy. You must add energy to climb.
Elliptical orbits. Launch sideways a little faster than the circular speed and the satellite swings out to a farthest
point, then falls back. It is fastest at the closest point and slowest at the farthest point. Both L and E stay constant.
Escape. If the total energy E = K + U is zero or more, the satellite never comes back. Set ½mv² − GMm/r = 0:
vescape = √(2GM / r) = √2 × vcircular
Constants used on this page: G = 6.67 × 10⁻¹¹ N·m²/kg², Earth's mass M = 5.97 × 10²⁴ kg, Earth's radius
RE = 6.37 × 10⁶ m, so GM = 3.98 × 10¹⁴ m³/s².
Worked: speed and period of the space station
Read the steps as text
The station orbits 420 km above the ground in a circle. Find its speed and the time for one orbit.
Find r from Earth's center: r = RE + 420 km = 6.37 × 10⁶ m + 0.42 × 10⁶ m = 6.79 × 10⁶ m.
r in the gravity law is center to center, not the height above the ground.
Gravity supplies the centripetal force: GMm/r² = mv²/r, so v = √(GM/r).
the only force is gravity, and it points to the center of the circle.
v = √(3.98 × 10¹⁴ / 6.79 × 10⁶) = √(5.86 × 10⁷) = 7.66 × 10³ m/s, about 7.7 km/s.
m cancelled, so we did not need the station's mass.
T = 2πr / v = 2π × 6.79 × 10⁶ m / 7660 m/s = 5.57 × 10³ s ≈ 93 minutes.
one lap is the circumference 2πr at constant speed. That is about 15.5 orbits per day.
Energy per kilogram: K/m = v²/2 = 2.93 × 10⁷ J/kg, U/m = −GM/r = −5.86 × 10⁷ J/kg, E/m = −2.93 × 10⁷ J/kg.
check the rule E = −K for a circular orbit: it works.
Lab: launch a satellite
Pick a height above the ground and a sideways launch speed. Time runs in hours. The bars show K, U and
E for each kilogram of satellite (in MJ per kg). Watch the purple E bar: it never changes, while K and U trade off.
Try this: at 400 km the circular speed is about 7.67 km/s. Try 7.67 (circle), 9.0 (ellipse: watch the speed
drop as r grows), 10.9 (escape: E becomes positive) and 6.0 (crash). Then try 35 800 km at 3.07 km/s: a 24 hour orbit.
Examples
Example 1: the 24 hour orbit basic
A weather satellite must orbit once every 24 h (86 400 s) so it stays above the same spot. Find the radius of its circular orbit and its speed.
Show solutionRead the steps as text
From T = 2π√(r³/GM): r³ = GM T² / (4π²) = 3.98 × 10¹⁴ × (86 400)² / (4π²) = 7.53 × 10²².
Square both sides of the period equation and solve for r³.
r = (7.53 × 10²²)1/3 = 4.22 × 10⁷ m, about 6.6 Earth radii (altitude about 3.6 × 10⁷ m).
Take the cube root.
v = 2πr / T = 2π × 4.22 × 10⁷ / 86 400 = 3.07 × 10³ m/s.
Much slower than the station: higher orbits are slower.
Example 2: raising an orbit medium
A 1000 kg satellite moves from a circular orbit with r = 7.0 × 10⁶ m to one with r = 1.4 × 10⁷ m. (a) Find its speed in each
orbit. (b) How much energy must the rocket add to the satellite-Earth system?
(b) E₁ = −GMm/(2r₁) = −3.98 × 10¹⁴ × 1000 / (1.4 × 10⁷) = −2.84 × 10¹⁰ J. E₂ = −GMm/(2r₂) = −1.42 × 10¹⁰ J.
Circular orbit energy E = −GMm/(2r).
Energy added = E₂ − E₁ = +1.42 × 10¹⁰ J. The higher orbit has less kinetic energy but much more
potential energy, so the total goes up. The satellite ends up slower even though energy was added.
Example 3: an elliptical orbit AP
A satellite's closest point is 7.0 × 10⁶ m from Earth's center, where its speed is 8.5 × 10³ m/s. Its farthest point is
1.22 × 10⁷ m from the center. (a) Find its speed at the farthest point. (b) Check with energy conservation.
Show solutionRead the steps as text
(a) Gravity makes no torque about Earth's center, so L is constant. At both points v ⟂ r, so r₁v₁ = r₂v₂.
L = mvr when the velocity is perpendicular to r; m cancels.
(b) Energy per kg at the close point: ½(8500)² − 3.98 × 10¹⁴/7.0 × 10⁶ = 3.61 × 10⁷ − 5.69 × 10⁷ = −2.08 × 10⁷ J/kg.
E/m = ½v² − GM/r.
At the far point: ½(4880)² − 3.98 × 10¹⁴/1.22 × 10⁷ = 1.19 × 10⁷ − 3.26 × 10⁷ = −2.07 × 10⁷ J/kg.
The same within rounding, so both laws agree. E is negative: the satellite is bound.
Example 4: escaping from orbit AP
A probe is in the same circular orbit as the station (r = 6.79 × 10⁶ m, v = 7.66 × 10³ m/s). How much faster must it go
to escape from Earth completely? Explain using energy.
Show solutionRead the steps as text
To escape, the total energy of the probe-Earth system must reach zero: ½mv² − GMm/r = 0.
With E = 0 the probe can just reach "infinitely far" (U = 0) with zero speed left.
vesc = √(2GM/r) = √(2 × 3.98 × 10¹⁴ / 6.79 × 10⁶) = 1.08 × 10⁴ m/s.
This is √2 times the circular speed: 1.414 × 7.66 = 10.8 km/s.
Extra speed needed = 10.8 − 7.66 ≈ 3.2 km/s. Starting from orbit saves a lot compared with starting
from the ground (11.2 km/s from the surface, before air drag).
Practice (AP style)
1. Satellite B orbits Earth in a circle of radius 4r. Satellite A orbits in a circle of radius r. What is vB / vA?
Show answer
B. v = √(GM/r). Four times the radius means √(1/4) = ½ the speed.
2. For the same two satellites, what is TB / TA?
Show answer
C. T ∝ r3/2. 43/2 = 8. Check: B travels a path 4 times longer at half the speed, so it takes 8 times as long.
3. A comet moves on a long elliptical orbit around the Sun. Consider the comet-Sun system. Which quantities stay constant during one orbit?
Show answer
C. Gravity points to the Sun, so it makes no torque about the Sun: L is constant. Gravity is internal and conservative, so K + U is constant. K and U each change; L stays constant because v grows when r shrinks.
4. A satellite's farthest point from Earth's center is 3 times its closest distance. At the closest point its speed is v. What is its speed at the farthest point?
Show answer
A. At both points v ⟂ r, so conservation of angular momentum gives r v = (3r) vfar, so vfar = v/3.
5. A satellite in a circular orbit has kinetic energy 1.5 × 10¹⁰ J. What are the gravitational potential energy U of the satellite-Earth system and the total energy E?
Show answer
B. For a circular orbit K = GMm/(2r) and U = −GMm/r = −2K. E = K + U = −K = −1.5 × 10¹⁰ J. (Choice A would mean the satellite is just escaping, not in orbit.)
6. A spacecraft is in a circular orbit with speed vc. Its engine fires briefly forward along its motion. What is the smallest new speed that lets it escape Earth?
Show answer
B. Escape needs ½mv² = GMm/r, so v = √(2GM/r) = √2 × √(GM/r) = √2 vc. 11.2 km/s is the escape speed from the ground, not from this orbit.
7. Two satellites share the same circular orbit. Satellite Q has twice the mass of satellite P. Which statement is true?
Show answer
C. v = √(GM/r) does not contain the satellite's mass, so speed and period are the same. But K, U and E are all proportional to m, so Q's E = −GMm/(2r) is twice as negative.
8. (Free response, translation between representations) A satellite is moved slowly through a series of larger and larger circular orbits. (a) Describe how K, U and E of the satellite-Earth system change as r increases, and sketch in words the shape of each graph against r. (b) A student says "the satellite needs more energy to orbit higher, but it moves more slowly, so that is impossible." Resolve the paradox.
Show answer
(a) K = GMm/(2r): positive, falls off as 1/r (a curve that drops quickly then flattens toward zero). U = −GMm/r: negative,
twice as far from zero as K, rising toward zero as r grows. E = −GMm/(2r): negative, the mirror image of K below the axis,
rising toward zero. At every r, E = −K and U = −2K.
(b) The energy added goes into potential energy. U rises by twice as much as K drops (ΔU = −2ΔK), so the total E rises
even though K falls. A higher orbit has more total energy and less kinetic energy, and there is no contradiction.
Free response (Mathematical routines). A weather satellite moves in a circular orbit of radius r around Earth (GM = 3.98 × 10¹⁴ m³/s²).
(a) Starting from Newton's second law, derive an expression for the period T in terms of r and GM.
(b) Calculate T for r = 4.22 × 10⁷ m. Give it in hours.
(c) Calculate the satellite's speed.
(d) Why does a satellite at this radius seem to hang still above one spot on the equator?
Show answer
(a) GMm/r² = mv²/r gives v = √(GM/r). One lap: T = 2πr / v = 2π√(r³/GM).
(c) v = √(3.98 × 10¹⁴ / 4.22 × 10⁷) ≈ 3.07 × 10³ m/s (about 3.1 km/s).
(d) Its period matches one turn of Earth (about a day), so moving in the same direction it stays over the same spot.
Point guide (4 points): 1 for gravity = centripetal force; 1 for T = 2π√(r³/GM); 1 for T ≈ 24 h; 1 for v ≈ 3.1 km/s.
Free response (Experimental design and analysis). Astronomers measure the orbit radius r and period T of four moons of Jupiter:
Io 4.22 × 10⁸ m, 1.77 days; Europa 6.71 × 10⁸ m, 3.55 days; Ganymede 1.07 × 10⁹ m, 7.15 days; Callisto 1.88 × 10⁹ m, 16.7 days.
(a) What should be graphed to get a straight line? What should the line look like if the orbits obey T² = 4π²r³/(GM)?
(b) Using the first and last moons, find the slope and Jupiter's mass (G = 6.67 × 10⁻¹¹ N·m²/kg²).
(c) Why don't you need the moons' own masses?
Show answer
(a) Plot T² (vertical, in s²) against r³ (horizontal, in m³). It should be a straight line through the origin with slope 4π²/(GM).
(c) In GMm/r² = mv²/r the moon's mass m cancels. The orbit depends only on the central mass.
Point guide (4 points): 1 for T² against r³; 1 for converting days to seconds; 1 for slope and M ≈ 1.9 × 10²⁷ kg; 1 for m cancelling.
Free response (Qualitative / quantitative translation). A satellite moves from a circular orbit of radius r to a circular orbit of radius 2r.
(a) Without numbers, say whether each of these goes up, goes down or stays the same: speed, kinetic energy,
gravitational potential energy, total energy of the satellite-Earth system.
(b) Write each quantity in the new orbit as a multiple of its old value.
(c) The satellite has m = 500 kg and r = 7.0 × 10⁶ m (GM = 3.98 × 10¹⁴ m³/s²). How much energy must the engines add?
Show answer
(a) Speed down, kinetic energy down, potential energy up (less negative), total energy up.
(b) v = √(GM/r), so v becomes v/√2 ≈ 0.71v. K = GMm/(2r) halves. U = −GMm/r becomes −GMm/(2r): half as negative, so it goes up.
E = −GMm/(2r) becomes −GMm/(4r): half as negative, so it goes up.