Gravitational Fields and Orbits
Gravitational Fields and Orbits
A gravitational field is a region of space where a mass experiences a force due to the gravitational attraction of another mass. Gravity is always attractive and acts on all objects with mass.
Newton's Law of Universal Gravitation
F = −GMm/r²
where:
- F = gravitational force (N)
- G = gravitational constant = 6.674 × 10⁻¹¹ N m² kg⁻²
- M, m = the two masses (kg)
- r = distance between their centres (m)
- The negative sign indicates the force is attractive
This is an inverse square law: doubling the distance reduces the force to one quarter.
Gravitational Field Strength
Gravitational field strength, g, is the force per unit mass experienced by a small test mass placed in the field:
g = F/m = GM/r² (for a point or spherical mass)
At the Earth's surface: g ≈ 9.81 N kg⁻¹ (equivalent to 9.81 m s⁻²)
Uniform field: Near the Earth's surface, g is approximately constant — field lines are parallel and equally spaced.
Radial field: Far from the surface, g decreases with r² — field lines point radially inward, converging toward the centre.
Gravitational Potential
Gravitational potential, V, at a point is the work done per unit mass in bringing a small test mass from infinity to that point:
V = −GM/r
Key points:
- V is always negative (work is done by the field as mass moves from infinity inward)
- V = 0 at infinity
- V becomes more negative closer to the mass (a deeper "potential well")
Gravitational Potential Energy
E_p = mV = −GMm/r
The change in potential energy between two points: ΔE_p = mΔV
Escape Velocity
The escape velocity is the minimum speed needed for an object to escape a gravitational field (reach infinity with zero kinetic energy).
Setting ½mv² = GMm/r:
v_escape = √(2GM/r) = √(2gr)
For Earth: v_escape = √(2 × 9.81 × 6.371 × 10⁶) = 11,200 m s⁻¹ ≈ 11.2 km s⁻¹
Orbital Motion
For a satellite in a circular orbit, gravity provides the centripetal force:
GMm/r² = mv²/r → v = √(GM/r)
The orbital speed decreases with increasing orbital radius.
Orbital period: T = 2πr/v, so:
T² = 4π²r³/(GM)
This is Kepler's Third Law: T² ∝ r³ for objects orbiting the same central mass.
Worked Example
Find the orbital period of the ISS at altitude 408 km above Earth (M_E = 5.97 × 10²⁴ kg, R_E = 6.371 × 10⁶ m).
r = R_E + h = 6.371 × 10⁶ + 408 × 10³ = 6.779 × 10⁶ m
T² = 4π²r³/(GM) = 4π² × (6.779 × 10⁶)³ / (6.674 × 10⁻¹¹ × 5.97 × 10²⁴)
T² = 4π² × 3.114 × 10²⁰ / (3.983 × 10¹⁴) = 3.086 × 10⁷
T = 5555 s = 92.6 minutes
Kepler's Laws
1. First Law: Planets move in elliptical orbits with the Sun at one focus
2. Second Law: A line joining a planet and the Sun sweeps out equal areas in equal times (conservation of angular momentum)
3. Third Law: T² ∝ r³ — the square of the orbital period is proportional to the cube of the semi-major axis
Geostationary Orbits
A geostationary satellite has:
- Period T = 24 hours (matches Earth's rotation)
- Orbits above the equator
- Orbits west to east (same direction as Earth's rotation)
- Appears stationary from the ground
Using T² = 4π²r³/(GM): r = 42,200 km from Earth's centre (altitude ≈ 35,800 km)
Applications: Communications satellites, weather monitoring, TV broadcasting
Energy of an Orbiting Satellite
For a circular orbit of radius r:
- Kinetic energy: E_k = ½mv² = GMm/(2r)
- Potential energy: E_p = −GMm/r
- Total energy: E_total = E_k + E_p = −GMm/(2r)
The total energy is negative (the satellite is bound). Note that E_total = −E_k = E_p/2.
To move to a higher orbit: energy must be added (E_total becomes less negative). The satellite slows down (lower v) but gains potential energy. This is why boosting a spacecraft into a higher orbit paradoxically results in a lower orbital speed.
Gravitational Field of Multiple Masses
At a point between two masses, gravitational field strengths are added as vectors. There exists a neutral point where the resultant field is zero (the fields from each mass cancel).
For masses M and m separated by distance d, the neutral point is at distance x from M where:
GM/x² = Gm/(d − x)² → x/(d − x) = √(M/m)