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Many students expect a heavier satellite to need more speed, or to take longer to go round. It does not. Here is why, a worked check with real numbers, and the one case where the orbiting body's mass starts to matter.
Does a satellite's period depend on its mass?
No, as long as the satellite is much lighter than the planet, which is true of every artificial satellite. The standard result, in OpenStax University Physics, section 13.4, comes from one equation.
Gravity pulls the satellite towards the planet's centre with force GMm/r². To move in a circle of radius r at speed v, the satellite needs a centripetal force mv²/r. Set them equal:
GMm/r² = mv²/r
The satellite's mass m appears on both sides and cancels.
Orbital speed: v = √(GM/r).
Period, one full circle 2πr divided by v: T = 2π√(r³/GM).
OpenStax puts it plainly: orbital velocity depends "only upon the distance from the center of the planet, and not upon the mass of the object being acted upon" (OpenStax, section 13.4). For the full derivation, see deriving orbital velocity and the time period of a satellite.
Why does the satellite's mass cancel out?
A heavier satellite feels a bigger pull, but needs an equally bigger force to turn its path, so its acceleration is the same. This is Galileo's falling-bodies result again.
In the final minutes of the third Apollo 15 moonwalk, commander David Scott dropped a 1.32 kg aluminium geological hammer and a falcon feather from about 1.6 m above the lunar surface, and the two struck the ground at the same time (NASA, Apollo 15 Hammer-Feather Drop). With almost no air, mass made no difference to how they fell.
An orbit is the same kind of fall, with enough sideways speed that the ground curves away beneath the satellite. It is also why astronauts float, as we explain in why astronauts feel weightless in orbit.
Worked example: a space station versus a 1 kg satellite
NASA's Earth fact sheet gives Earth's gravitational parameter as GM = 3.986 × 10¹⁴ m³/s² and its mean radius as 6,371 km (NASA Earth Fact Sheet). Take an orbit 400 km above the surface.
Radius from Earth's centre: r = 6,371 + 400 = 6,771 km = 6.771 × 10⁶ m.
Speed: v = √(3.986 × 10¹⁴ / 6.771 × 10⁶) ≈ 7,670 m/s, about 7.7 km/s.
Period: T = 2πr / v ≈ 5,540 s, about 92 minutes.
Notice that no satellite mass was used anywhere.
OpenStax works the same 400 km case and gets 5.55 × 10³ s, "just over 90 minutes" (OpenStax, Example 13.9). NASA lists the International Space Station's mass as 419,725 kg and says it orbits Earth about every 90 minutes (NASA, Space Station Facts and Figures). A 1 kg satellite at the same height would keep pace with it, lap after lap.
When does the orbiting body's mass matter?
Only when it is not tiny compared with the body it orbits. For two bodies circling each other, the exact period is T = 2π√(a³/G(M + m)), which reduces to the simple formula when m is negligible (Orbital period, Wikipedia).
Space station: 419,725 kg (NASA, Space Station Facts and Figures) against Earth's 5.97 × 10²⁴ kg (NASA Earth Fact Sheet) is a ratio of about 7 × 10⁻²⁰. The correction to a 92-minute period is far less than a trillionth of a second.
The Moon: its mass is 0.0123 of Earth's, its orbit's semi-major axis is 384,400 km, and its measured orbital period is 27.32 days (NASA Moon Fact Sheet). Ignoring the Moon's mass, T = 2π√(a³/GM) gives about 27.45 days. Including it, the factor 1/√(1.0123) brings this to about 27.28 days, much closer.
So for exam satellites, use the simple formula; for binary stars or Earth and Moon, use M + m.
How to use this in JEE and board problems
Questions often give the satellite's mass anyway. Check before you calculate:
Is the question asking for speed, period or angular speed? Ignore the satellite's mass.
Is it asking for kinetic energy, potential energy, total energy, angular momentum or the gravitational force? Keep the mass. OpenStax gives the total energy of a circular orbit as E = −GMm/2r, which has m in it (OpenStax, section 13.4).
Comparing two orbits around the same planet? Use ratios: T scales as r^(3/2) and v as 1/√r, so G and M never need numbers.
A classic trap: satellites of mass m and 2m in the same orbit have the same speed and period, but the heavier one has twice the kinetic energy, and its potential and total energies are twice as negative.
For practice comparing periods, see Kepler's third law period-ratio problems. If you keep choosing the wrong formula, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer, so you see where mass belongs. For orbits you would rather watch than read about, try the Fermi Simulation Lab.
