AP Physics 1 · Lesson 3

Orbits and satellites

Know Know it

In a circular orbit, gravity is the centripetal force: GMm/r2=mv2/rGMm/r^2 = mv^2/r, so v=GM/rv = \sqrt{GM/r}, where M is the central body’s mass and r is measured from its center. The satellite’s own mass cancels. The period follows from T=2πr/vT = 2\pi r/v, which gives T2=4π2r3/(GM)T^2 = 4\pi^2r^3/(GM).

In an elliptical orbit, gravity always points at the central body, so it exerts no torque about it, and the orbiting object’s angular momentum about the body is constant. With U=−GMm/rU = -GMm/r, the total mechanical energy is constant too. Both say the same thing: the object moves fastest at its closest point and slowest at its farthest.

Worked example Easy

A moon moves in a circular orbit around a planet. What provides the centripetal force that keeps it on its circle?

  1. The planet’s gravitational pull on the moonAnswer
  2. An outward force that balances gravity
  3. The moon’s own speed
  4. No force, because the moon is weightless in orbit

How to solve it

  1. Circular motion needs a net force toward the center.
  2. The only force on the moon is the planet’s gravity, and it points toward the planet’s center.

Why each choice is right or wrong

  • A. Correct. The planet’s gravity pulls the moon toward the planet’s center, which is exactly the inward force circular motion needs.
  • B. Incorrect. An outward force balancing gravity would leave no net inward force, and the moon would travel in a straight line.
  • C. Incorrect. The moon’s speed is not a force; it describes the motion that the inward force keeps bending.
  • D. Incorrect. The moon is not weightless: gravity acts on it the whole time, and that pull is what curves its path.

Trap Floating in orbit is not the absence of gravity

Astronauts in an orbiting station feel weightless because they and the station fall around Earth together with the same acceleration. At a typical station height, gravity is still about 90 percent as strong as at the surface.

Explain Explain it

Task verbs: calculate, explain

At the closest and farthest points of an elliptical orbit, the velocity is perpendicular to the line to the central body, so L=mvrL = mvr there, and v1r1=v2r2v_1r_1 = v_2r_2. An object twice as far away moves half as fast at those points.

To escape, an object needs total energy of at least zero: 12mv2−GMm/R≥0\frac{1}{2}mv^2 - GMm/R \geq 0, so the escape speed from a surface of radius R is 2GM/R\sqrt{2GM/R}, which is 2\sqrt{2} times the speed of an orbit skimming that surface.

Worked example Medium

A probe orbits Mars in a circle of radius 4.0×1064.0 \times 10^{6} m, measured from Mars’s center. Mars has a mass of 6.42×10236.42 \times 10^{23} kg, and G=6.67×10−11G = 6.67 \times 10^{-11} N·m²/kg². What is the probe’s orbital speed?

  1. 2.3×1032.3 \times 10^{3} m/s
  2. 3.3×1033.3 \times 10^{3} m/sAnswer
  3. 4.6×1034.6 \times 10^{3} m/s
  4. 1.1×1071.1 \times 10^{7} m/s

How to solve it

  1. Gravity is the centripetal force: v=GM/rv = \sqrt{GM/r}.
  2. GM/r=(6.67×10−11)(6.42×1023)/(4.0×106)≈1.07×107GM/r = (6.67 \times 10^{-11})(6.42 \times 10^{23})/(4.0 \times 10^{6}) \approx 1.07 \times 10^{7} m²/s².
  3. v≈3.3×103v \approx 3.3 \times 10^{3} m/s.

Why each choice is right or wrong

  • A. Incorrect. 2.3×1032.3 \times 10^{3} m/s uses twice the radius in GM/r\sqrt{GM/r}.
  • B. Correct. v=GM/r=(6.67×10−11)(6.42×1023)/(4.0×106)≈3.3×103v = \sqrt{GM/r} = \sqrt{(6.67 \times 10^{-11})(6.42 \times 10^{23})/(4.0 \times 10^{6})} \approx 3.3 \times 10^{3} m/s.
  • C. Incorrect. 4.6×1034.6 \times 10^{3} m/s is 2GM/r\sqrt{2GM/r}, the escape speed from that distance, not the circular orbit speed.
  • D. Incorrect. 1.1×1071.1 \times 10^{7} m/s is GM/rGM/r without the square root; that value is v2v^2, in m²/s².

Calculate Your turn

A small moon moves in an elliptical orbit around a planet. At its closest point it is 2.0×1082.0 \times 10^{8} m from the planet’s center and moves at 3.0 km/s; at its farthest point it is 5.0×1085.0 \times 10^{8} m away. At both points its velocity is perpendicular to the line to the planet. Calculate the moon’s speed at its farthest point, and explain why its angular momentum about the planet stays constant.

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Compare with a model answer

A full answer does this

  • Gives the speed at the farthest point as 1.2 km/s
  • Says the angular momentum stays constant because the planet’s gravity exerts no torque on the moon about the planet

Model answer

At both points the velocity is perpendicular to the radius, so L=mvrL = mvr and v1r1=v2r2v_1r_1 = v_2r_2: v2=(3.0 km/s)(2.0×108)/(5.0×108)=1.2v_2 = (3.0 \text{ km/s})(2.0 \times 10^{8})/(5.0 \times 10^{8}) = 1.2 km/s. The angular momentum stays constant because the only force on the moon, the planet’s gravity, always points toward the planet’s center; a force along the line to the point has no lever arm, so it exerts no torque about the planet.

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