Satellite in Circular Orbit

For a satellite in a stable circular orbit, gravity provides the centripetal force. Setting Newton's law of gravitation equal to the centripetal force formula:

$\dfrac{GMm}{r^{2}} = \dfrac{mv^{2}}{r}$

The satellite's mass $m$ cancels (just like for free fall on a planet's surface!) and we can solve for the orbital radius:

$r = \dfrac{GM}{v^{2}}$

Or, given the radius, the orbital speed must be:

$v = \sqrt{\dfrac{GM}{r}}$

These are the equations that govern the entire space-flight era. Pick any speed, and there's exactly one circular-orbit radius that goes with it. Pick any radius, and there's exactly one speed.

Step-by-Step Solution

Given: Orbital speed $v = 8000\;\text{m/s}$, $G = 6.674 \times 10^{-11}\;\text{N}\cdot\text{m}^{2}/\text{kg}^{2}$, Earth mass $M = 5.972 \times 10^{24}\;\text{kg}$, Earth radius $R_\oplus = 6.371 \times 10^{6}\;\text{m}$.

Find: Orbital radius, altitude above surface, and orbital period.

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Step 1 — Compute $GM$ for Earth.

$GM = (6.674 \times 10^{-11})(5.972 \times 10^{24})$

$\approx 3.986 \times 10^{14}\;\text{m}^{3}/\text{s}^{2}$

(This product is sometimes called the standard gravitational parameter $\mu$ — for Earth it's a famous constant in astrodynamics.)

Step 2 — Solve for the orbital radius.

$r = \dfrac{GM}{v^{2}} = \dfrac{3.986 \times 10^{14}}{(8000)^{2}}$

$= \dfrac{3.986 \times 10^{14}}{6.4 \times 10^{7}}$

$\approx 6.228 \times 10^{6}\;\text{m}$

That's about 6228 km from the center of Earth.

Step 3 — Find the altitude above Earth's surface.

$h = r - R_\oplus = 6228 - 6371 = -143\;\text{km}$

Hmm — that's below the surface! What's going on? The answer is that 8000 m/s is just below the actual minimum orbital speed for the lowest possible orbit (the surface itself, where $v_{\text{circ}} = \sqrt{gR_\oplus} \approx 7905\;\text{m/s}$).

For a more realistic Low Earth Orbit (LEO), satellites typically travel at about 7600–7800 m/s at altitudes of 300–600 km. The slower speed corresponds to a larger radius, by the inverse-square law.

Step 4 — Compute the period.

$T = \dfrac{2\pi r}{v} = \dfrac{2\pi \times 6.228 \times 10^{6}}{8000}$

$= \dfrac{3.913 \times 10^{7}}{8000}$

$\approx 4891\;\text{s}$

Convert to minutes:

$T \approx \dfrac{4891}{60} \approx 81.5\;\text{minutes}$

A satellite at this radius circles the Earth in about 81.5 minutes — comparable to the International Space Station's ~92-minute orbit at 400 km altitude.

Step 5 — How does period scale with altitude?

Combining $T = 2\pi r/v$ with $v = \sqrt{GM/r}$ gives Kepler's Third Law:

$T^{2} = \dfrac{4\pi^{2}}{GM}\,r^{3}$

So $T \propto r^{3/2}$ — the higher the orbit, the slower the satellite moves and the longer the orbit takes. Geostationary satellites (24-hour period) are at 42{,}164 km radius — more than 6× higher than LEO.

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Answer:

• Orbital radius: $r = GM/v^{2} \approx 6.228 \times 10^{6}\;\text{m}$ (≈ 6228 km from Earth's center)
• Period: $\boxed{T = 2\pi r/v \approx 4891\;\text{s} \approx 81.5\;\text{min}}$

The radius implies a sub-surface orbit, which means 8000 m/s is slightly faster than what's required for a real LEO. The actual ISS orbit at 7670 m/s gives an orbital radius of about 6770 km — about 400 km altitude above the surface — with a 92-minute period.

Try It

• Adjust the orbital speed.
• The visualization shows the satellite circling Earth at the computed radius.
• The HUD reports the period and altitude above the surface.
• Try 3070 m/s — that gives geostationary altitude (42{,}164 km radius, 24-hour period).
• Try higher speeds — the orbit shrinks toward the surface and the period decreases.