Gravitational field of a point mass
Section: Gravitational Fields | Syllabus: Cambridge AS Level Physics 9702
Field Strength for a Point Mass Using Newton's Law of Gravitation (F = GMm/r^2) and the definition g = F/m, we can calculate the field strength created by a single large mass M. mg = GMm/r^2 g = GM/r^2 This shows that field strength follows the inverse square law relative to the distance from the centre of the mass.
Determining Planetary Mass If we know the orbital radius (r) and period (T) of a moon, we can calculate the mass of the planet it orbits. GMm/r^2 = mrω^2 M = r^3ω^2/G where ω = 2π/T Worked Example: Mass of Jupiter Using the moon **Io** (r = 4.22 × 10^8 m, T = 42 hours): Solution ω = 2π/42 × 3600 = 4.16 × 10^-5 rad s^-1 M_Jupiter = (4.22 × 10^8)^3 × (4.16 × 10^-5)^26.67 × 10^-11 1.9 × 10^27 kg.
Variations in Field Strength In reality, Earth's g is not perfectly constant over the surface due to: Oblate Spheroid Shape: Earth is slightly flattened at the poles and bulges at the Equator. Since the poles are closer to the centre, g is slightly higher there.
Density Variations: Differences in the density of underlying rocks. The Neutral Point Between Earth and the Moon, there is a **neutral point** where the gravitational fields of both bodies are equal in magnitude but opposite in direction.
As Earth is more massive, this point is much closer to the Moon. Figure 13.7: Neutral Point X Show Earth and Moon. At point X (near the Moon), the vector sum of g_Earth and g_Moon is zero. Geostationary Satellites For communication (like GSAT-11), receiving dishes must point to a fixed spot.
A **geostationary orbit** achieves this by requiring: Orbit from West to East (same as Earth). Orbit directly above the Equator . Period of exactly 24 hours .
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