13.2 Gravitational force between point masses
- Syllabus
- 9702–2028–2029
- Topic
- 13.2
- Level
- A2
Learning objectives
13.2.1That, for a point outside a uniform sphere, the mass of the sphere may be• understand that, for a point outside a uniform sphere, the mass of the sphere may be considered to be a point mass at its centre13.2.2Newton’s law of gravitation F = Gm1m2 / r2 for the force between two point• recall and use Newton’s law of gravitation F = Gm1m2 / r2 for the force between two point masses13.2.3Circular orbits in gravitational fields by relating the gravitational force• analyse circular orbits in gravitational fields by relating the gravitational force to the centripetal acceleration it causes13.2.4A satellite in a geostationary orbit remains at the same point above the• understand that a satellite in a geostationary orbit remains at the same point above the Earth’s surface, with an orbital period of 24 hours, orbiting from west to east, directly above the Equator
Outside a uniform sphere, use its total mass at the centre
For a point outside a uniform sphere, the gravitational force and field are exactly the same as if the sphere's total mass were concentrated at a point at its centre.
ForheighthaboveasphereofradiusR:r=R+h
| Situation | Point-mass use | Distance r |
|---|---|---|
| outside uniform spherical planet/star | valid | centre to external point |
| two non-overlapping uniform spheres | valid for their mutual force | centre-to-centre separation |
| very distant compact object | approximate when size ≪ separation | centre-to-centre separation |
| point inside the sphere | this external result is not valid | requires interior mass analysis |
A satellite 400 km above a planet of radius 6400 km has orbital radius r = 6800 km = 6.80 × 10⁶ m. Using 400 km as r would overestimate gravity severely.
The source may be represented at its centre; the external test object is not moved there. The theorem's external condition must be checked before using point-mass equations.
Newton's law gives the attraction between two point masses
The gravitational force between two point masses is directly proportional to the product of their masses and inversely proportional to the square of their separation. It acts attractively along the line joining them.
F=Gm1m2/r2,G=6.67×10−11Nm2kg−2
Convert all quantities to SI, use centre-to-centre separation r, square the complete separation, and report the force magnitude with its attractive direction. The two masses experience equal-magnitude opposite forces.
For m₁ = 3.0 kg, m₂ = 5.0 kg and r = 2.0 m, F = (6.67 × 10⁻¹¹)(3.0)(5.0)/(2.0)² = 2.50 × 10⁻¹⁰ N on each mass, directed toward the other.
Doubling either mass doubles F; doubling r makes F one quarter. The law is inverse-square, not inverse-distance, and Newton's third law prevents different force magnitudes on the two masses.
Gravity supplies the centripetal resultant in a circular orbit
For a satellite of mass m in a circular orbit of centre radius r around a spherical body of mass M, gravitational attraction is the inward resultant force. Velocity is tangential and gravity is perpendicular to it.
GMm/r2=mv2/r=mrω2
v2=GM/r,sov=√(GM/r)
usingv=2πr/T:T2=4π2r3/(GM)andr3=GMT2/(4π2)
| Change for the same central mass M | Circular-orbit consequence |
|---|---|
| larger r | smaller v, smaller acceleration, longer T |
| smaller r | larger v, larger acceleration, shorter T |
| different satellite mass m at same r | same v and T because m cancels |
Do not add a separate centripetal force to gravity: gravity is the centripetal resultant. If altitude h is given, use r = planet radius + h. The object is not force-free; it is continuously falling around the body.
A geostationary orbit satisfies four simultaneous conditions
| Required condition | Why it is needed |
|---|---|
| circular orbit | constant radius and angular speed |
| directly above the Equator | orbital plane matches Earth's equatorial rotation |
| west to east | same direction as Earth's rotation |
| period 24 h | same angular speed as Earth |
When all four conditions hold, the satellite remains above the same point/longitude on Earth's surface and appears stationary to an Earth-fixed observer.
A 24-hour polar or tilted orbit crosses different latitudes; a 24-hour westward orbit moves relative to Earth; an elliptical synchronous orbit changes radius and apparent position. Equal period alone is therefore insufficient.
Because its sky position is fixed, a communications dish can point continuously at one geostationary satellite without tracking it across the sky.
Geostationary is Earth-specific wording here: 24 h matches Earth's rotation. A stationary-looking orbit around another planet must match that planet's own rotation period while retaining equatorial, same-direction and circular conditions.