Low Earth orbit: speed, period, field strength and true weight

SPH4U Grade 12 Physics · Gravitational, Electric and Magnetic Fields

A mission-planning exercise casts the class as the flight dynamics team for a new Earth-observation mission. An imaging satellite of mass 1,250 kg has separated from its launcher into a circular orbit of radius 7,200 km measured from the centre of Earth — about 830 km above the surface, since Earth's mean radius is 6,371 km. On the tracking display the orbit draws itself as a clean ring over the rotating globe, a fresh ground track unrolling with each pass. Before the first imaging pass is scheduled, the orbit's basic numbers have to be checked by hand. 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.

Every number in this problem is editable — change any value below and the whole chain recalculates.

Given

m = 1,250 kg — Imaging satellite
r = 7,200 km — Orbital radius, from Earth's centre
M = 5.972e+24 kg — Mass of Earth

Determine

(a)the orbital speed
(b)the period of one lap
(c)Earth's gravitational field strength at that radius
(d)the gravitational force on the satellite

Step 1 of 4(a) · solve for Orbital velocity

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 7,200 km has to land in the formula as 7.20 × 10⁶ m.

Rearranged for v

v = \sqrt{\tfrac{GM}{r}}

M

r

Your values, in your units

v = \sqrt{\tfrac{\left( 6.67430e-11\ \text{m}^{3}\text{/(kg}{\cdot}\text{s}^{2}\text{)} \right) \, \left( 5.97200e+24\ \text{kg} \right)}{\left( 7{,}200\ \text{km} \right)}}

Converted to base units

v = \sqrt{\tfrac{\left( 6.67430e-11\ \text{m}^{3}\text{/(kg}{\cdot}\text{s}^{2}\text{)} \right) \, \left( 5.97200e+24\ \text{kg} \right)}{\left( 7{,}200{,}000\ \text{m} \right)}}

Answer

v = 7.4404\ \text{km/s}

in

Carried onward at full precision, not this rounded figure.

Open the Orbital Velocity solver →

Step 2 of 4(b) · solve for Period

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.

Rearranged for T

T = \frac{2\pi r}{v}

v7.4404 km/scarried from step 1

r

Your values, in your units

T = \frac{2\left( 3.14159 \right) \, \left( 7{,}200\ \text{km} \right)}{\left( 7{,}440.4\ \text{m/s} \right)}

Converted to base units

T = \frac{2\left( 3.14159 \right) \, \left( 7{,}200{,}000\ \text{m} \right)}{\left( 7{,}440.4\ \text{m/s} \right)}

Answer

T = 101.34\ \text{min}

in

Carried onward at full precision, not this rounded figure.

Open the Speed in Circular Motion (v = 2πr/T) solver →

Step 3 of 4(c) · solve for Centripetal acceleration

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.

Rearranged for ac

a_c = \frac{v^2}{r}

v7.4404 km/scarried from step 1

r

Your values, in your units

a_c = \frac{\left( 7{,}440.4\ \text{m/s} \right)^2}{\left( 7{,}200\ \text{km} \right)}

Converted to base units

a_c = \frac{\left( 7{,}440.4\ \text{m/s} \right)^2}{\left( 7{,}200{,}000\ \text{m} \right)}

Answer

a_c = 7.6888\ \text{m/s}^{2}

in

Carried onward at full precision, not this rounded figure.

Open the Centripetal Acceleration (a = v²/r) solver →

Step 4 of 4(d) · solve for Force

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.

Rearranged for F

F = m a

m

a7.6888 m/s²carried from step 3

Your values, in your units

F = \left( 1{,}250\ \text{kg} \right) \, \left( 7.68883\ \text{m/s}^{2} \right)

Answer

F = 9.611\ \text{kN}

in

Carried onward at full precision, not this rounded figure.

Open the Newton's Second Law solver →

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 1,250 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 — 7,200 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 7,200 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.