Low Earth orbit: speed, period, field strength and true weight
SPH4U Grade 12 Physics · Gravitational, Electric and Magnetic Fields
An imaging satellite of mass 1250 kg is placed in a circular orbit of radius 7200 km measured from the centre of Earth — about 830 km above the surface, since Earth's mean radius is 6371 km. Taking Earth's mass as 5.972 × 10²⁴ kg, find the satellite's orbital speed, the time it takes to complete one lap, the strength of Earth's gravitational field at that radius, and the gravitational force Earth still exerts on the satellite.
A circular orbit has exactly one speed for a given radius — set gravity equal to mv²/r and the satellite's own mass cancels. Everything else in this chain is downstream of this number. Orbits are quoted in kilometres and G is written in metres, so the 7200 km has to land in the formula as 7.20 × 10⁶ m.
Carried onward at full precision, not this rounded figure.
One circumference at that speed. No new physics — just 2πr divided by the answer to step 1, with the radius back in metres so it matches the m/s underneath it. The answer is close to 101 minutes.
Carried onward at full precision, not this rounded figure.
v²/r is the satellite's centripetal acceleration, and since gravity is the only thing acting, it is also the field strength g at this radius. Two names, one number — that identity is the point of the step.
Carried onward at full precision, not this rounded figure.
Multiply by the satellite's mass and you have the gravitational force on it. Compare it with 12 258 N, what the same satellite weighs on the launch pad, before deciding whether 'weightless' was ever the right word.
Carried onward at full precision, not this rounded figure.
Why this order
Radius first, always. Orbital speed, period, field strength and force are four questions with one answer between them, and the bridge is r. Step 1 gets there by setting GMm/r² equal to mv²/r; the satellite's mass cancels on both sides, which is why a bolt and a bus share an orbit perfectly happily and why step 1 does not ask for 1250 kg. Step 4 is the first place the satellite's own mass is allowed in, and only because force, unlike acceleration, scales with what is being pushed.
Step 3 is the one worth arguing about. The answer is 7.69 m/s², which is 78% of the 9.81 m/s² you feel standing on the ground, and the satellite genuinely weighs 9.61 kN. Nothing is cancelling gravity. What astronauts experience is free fall: the station and everything in it accelerate toward Earth at that same 7.69 m/s² and simply keep missing, because they are also moving sideways at 7.44 km/s. Newton drew exactly this in De mundi systemate — a cannon on a mountain, fired harder and harder until the ball's fall curves away as fast as the Earth does. The other reliable slips are both about r. It is quoted in kilometres, as every orbit is, and it must reach the formula in metres — 7200 km is 7.20 × 10⁶ m, and a factor of a thousand here changes the orbital speed by a factor of thirty. It is also measured from Earth's centre, not from the ground: feeding in the 830 km of altitude instead of the 7200 km radius gives a speed nearly three times too high. And because r³ sits under the period while r² sits under the field strength, a 15% change in radius means very different things to the two answers, so r has to be carried at full precision rather than rounded early.
Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.