A geostationary satellite orbits Earth once every 86,400 s. Taking GM = 4.0×10¹⁴ m³/s² and Earth's radius as 6.4×10⁶ m, calculate the height of the orbit above Earth's surface, in kilometres.

Gravity Challenge #5 — Geostationary Orbits

The height of a
geostationary orbit.

Gravity Hard

Problem

A geostationary satellite circles Earth exactly once every 86,400 s, hovering over one spot on the equator. Calculate its height above Earth’s surface, in kilometres.

Orbital period\( T = 86{,}400\ \mathrm{s} \)
Earth’s GM\( 4.0 \times 10^{14}\ \mathrm{m^3/s^2} \)
Earth’s radius\( R_E = 6.4 \times 10^6\ \mathrm{m} \)
Answer inkilometres

Height means above the surface — the orbit radius is measured from Earth’s centre.

Hint: gravity supplies the centripetal force with \( \omega = \tfrac{2\pi}{T} \), which packs into \( r^3 = \tfrac{GMT^2}{4\pi^2} \). Cube-root for \( r \) — then remember what the question asked for.

Create a free account to solve

Submit your answer and unlock the full mark scheme.

Full mark schemes, not just answers
Earn XP and climb the leaderboard
Track your mastery, topic by topic

Create free account

Already have an account?

Finish a lesson to solve

Complete the Gravity lessons to unlock solving and earn XP from this challenge — or skip ahead with Physi+.

Continue Learning

Or unlock everything with Physi+

You solved this challenge

You can open the mark scheme below. Check again anytime — no extra XP.

Submit your answer

Solve it on paper first — then check yourself.

Your answer

Numeric answers allow ±1% tolerance for rounding.

Up next · Dynamics The 25-Joule Launch Keep the streak going — a fresh problem in Dynamics.

Mark schemeClick to open

The height of ageostationary orbit.

Problem

Submit your answer

The height of a
geostationary orbit.