Gravitational potential and orbits

Gravitational potential

The gravitational potential V_grav at a point is the work done per unit mass in bringing a small mass from infinity to that point. For a point mass M:

V_grav = -GM / r

It is negative everywhere because gravity is attractive: work must be done to move a mass out to infinity, where V = 0. The work done moving mass m between two points is ΔW = mΔV_grav.

Field and potential

As with electric fields, the field strength is the negative potential gradient: g = -ΔV / Δr. Equipotentials around a planet are concentric spheres; no work is done moving along one.

Orbits

For a satellite of mass m in a circular orbit of radius r, gravity provides the centripetal force:

GMm / r² = mv² / r, so v = sqrt(GM / r)

Using v = 2πr / T gives T² = 4π^2r3 / GM. The satellite's mass cancels. Lower orbits are faster and have shorter periods.

Geostationary orbits

A geostationary satellite orbits above the equator, in the same direction as the Earth's rotation, with a period of 24 hours. It stays above the same point, which is useful for communications. Its orbital radius is about 4.2 x 10⁷ m.

Worked example

Find the orbital speed of a satellite at radius 6.77 x 10⁶ m from the Earth's centre. (M = 5.97 x 10²⁴ kg)

1. v = sqrt(GM / r) = sqrt(6.67 x 10^-11 x 5.97 x 10²⁴ / 6.77 x 10⁶)
2. = sqrt(5.88 x 10⁷)

Answer: v = 7670 m/s (about 7.7 km/s)

Check yourself

1. Find the gravitational potential at the Earth's surface, in units of 10⁷ J/kg, to 3 s.f. (M = 5.97 x 10²⁴ kg, R = 6.37 x 10⁶ m)
   Show answer

   -6.25 x 10⁷ J/kg. V = -GM / r = -6.67 x 10^-11 x 5.97 x 10²⁴ / 6.37 x 10⁶ = -6.25 x 10⁷ J/kg.

2. How much energy is needed to raise a 1000 kg satellite from the Earth's surface (r = 6.37 x 10⁶ m) to r = 6.77 x 10⁶ m? (M = 5.97 x 10²⁴ kg) Answer in units of 10⁹ J, to 3 s.f.
   Show answer

   3.69 x 10⁹ J. ΔW = mΔV = 1000 x 3.98 x 10¹⁴ x (1/6.37 x 10⁶ - 1/6.77 x 10⁶) = 1000 x 3.98 x 10¹⁴ x 9.28 x 10^-9 = 3.69 x 10⁹ J.

3. Find the orbital speed of a satellite at r = 6.77 x 10⁶ m from the Earth's centre, in m/s, to 3 s.f. (M = 5.97 x 10²⁴ kg)
   Show answer

   7670 m/s. v = sqrt(GM / r) = sqrt(3.98 x 10¹⁴ / 6.77 x 10⁶) = 7670 m/s.

4. Find the period of that orbit (r = 6.77 x 10⁶ m, v = 7670 m/s), in minutes, to 3 s.f.
   Show answer

   92.4 min. T = 2πr / v = 2π x 6.77 x 10⁶ / 7669 = 5546 s = 92.4 minutes.

5. Find the orbital radius of a geostationary satellite (T = 24 h), in units of 10⁷ m, to 3 s.f. (M = 5.97 x 10²⁴ kg)
   Show answer

   4.22 x 10⁷ m. r³ = GMT² / 4π² = 3.98 x 10¹⁴ x (86400)² / 39.48 = 7.53 x 10²² m³, so r = 4.22 x 10⁷ m.

6. Why is gravitational potential always negative?
   • It is zero at infinity and work must be done to move a mass away to infinity
   • Gravity is a weak force
   • Masses can be negative
   • It is measured downwards
   Show answer

   It is zero at infinity and work must be done to move a mass away to infinity. Moving towards a mass, the potential falls below zero.

7. How does a satellite in a lower circular orbit compare with one in a higher orbit?
   • It moves slower and has a longer period
   • It moves faster and has a longer period
   • It moves faster and has a shorter period
   • It has the same speed
   Show answer

   It moves faster and has a shorter period. v = sqrt(GM / r), so a smaller r gives a larger v; T² is proportional to r³.

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Sources

• Pearson Edexcel International Advanced Subsidiary/Advanced Level in Physics specification (XPH11/YPH11), Issue 3
• Pearson qualifications: International Advanced Levels Physics (2018)
• Pearson Edexcel IAL Physics Unit 5 (WPH15) question paper, June 2023: list of data, formulae and relationships

Last updated 7 November 2026. Spotted a mistake? Tell us.