Orbits: Period, Speed and Kepler's Third Law · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Circular orbits and geostationary satellites

Science · Space Systems & Earth's History · ages 17-18
Name ______________________   Date ____________

  1. Two small satellites have circular orbits around the same planet. B's radius is four times A's. How does B's speed compare with A's?

    • It is twice A's speed.
    • It is four times A's speed.
    • It is half A's speed.

  2. Which set of conditions makes an Earth satellite geostationary?

    • Circular, equatorial, eastward; period equals Earth's sidereal day.
    • Circular, equatorial, westward; period equals Earth's sidereal day.
    • Circular, polar, northward at the equator; period equals Earth's sidereal day.

  3. For small satellites in circular orbits around the same central mass, plotting T² vertically against r³ horizontally gives a straight line through the origin.

    Circle one:   True   False

  4. For a small satellite in a circular orbit, which equation sets gravity equal to the required centripetal force?

    • GMm/r² = mv²/r²
    • GMm/r² = mv²/r
    • GMm/r² = mv/r

  5. Two small satellites have circular orbits around the same planet. B's orbital radius is nine times A's. By what factor is B's period longer?

    Answer: ______________

  6. A small satellite has circular orbital period T around a body of mass M. Which expression gives its orbital radius r?

    • r = √(GMT²/(4π²))
    • r = ∛(4π²T²/(GM))
    • r = ∛(GMT²/(4π²))

  7. Nina says a satellite with twice the mass needs twice the speed in the same circular orbit. Which step in the force equation corrects her claim?

    • The factor m cancels from both sides of GMm/r² = mv²/r.
    • The central mass M cancels from both sides of GMm/r² = mv²/r.
    • The radius r cancels completely, so the speed is independent of orbital radius.

  8. Calculate a geostationary satellite's height in km, to the nearest 100 km. Use GM = 3.986 × 10^14 m³/s², T = 86,164 s and Earth's radius R = 6,370 km. Use the small-satellite circular-orbit model.

    Answer: ______________

LightMySky · lightmysky.comW1-mt_PLKLykGF8Z-s1

Answer key

For grown-ups. Fold this page away before handing over the rest.

Circular orbits and geostationary satellites W1-mt_PLKLykGF8Z-s1

  1. It is half A's speed. · At fixed GM, v² = GM/r. Four times the radius gives one quarter of the speed squared, so B's speed is half A's.
  2. Circular, equatorial, eastward; period equals Earth's sidereal day. · The satellite must orbit eastward in a circular equatorial orbit with a period matching Earth's sidereal rotation period.
  3. True · T² = [4π²/(GM)]r³ has the form y = kx. With the same central mass, k is constant and the line passes through the origin.
  4. GMm/r² = mv²/r · The gravitational force is GMm/r². Circular motion requires an inward force mv²/r, supplied here by gravity.
  5. 27 · The period factor is √(9³) = √729 = 27.
  6. r = ∛(GMT²/(4π²)) · Rearranging T² = 4π²r³/(GM) gives r³ = GMT²/(4π²). Take the cube root to find r.
  7. The factor m cancels from both sides of GMm/r² = mv²/r. · The satellite mass m cancels. For the same central mass and circular radius, the required speed is √(GM/r), independent of the small satellite's mass.
  8. 35800 · The period gives r ≈ 42,164 km from Earth's centre. Subtract 6,370 km: h ≈ 35,794 km, which rounds to 35,800 km.
Worksheet · LightMySky