Example 1 · level 1
Find the acceleration due to gravity at a height of 3,200 km above Earth's surface (R = 6400 km, g = 9.8 m/s²), correct to 2 decimal places (in m/s²).
Show solution
- g′ = 9.8 ÷ 1.5² = 9.8 ÷ 2.25 = 4.36 m/s².
Answer: 4.36
CLASS 11–12 PHYSICS · GRAVITATION
Gravitation: g with height, orbital and escape speed
g at height h: g′ = g R²/(R + h)². Orbital speed at radius r: v = √(GM/r) = √(gR²/r). Escape speed ve = √(2gR). Use R(Earth) = 6400 km, g = 9.8 m/s².
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Worked examples
Example 2 · level 2
Find the orbital speed of a satellite orbiting 3,600 km above Earth's surface (R = 6400 km, g = 9.8 m/s²). Give the answer in km/s, correct to 2 decimal places.
Show solution
- v = √(gR²/r) = √(9.8 × (6.4 × 10⁶)² ÷ 10 × 10⁶)
- = 6.34 km/s.
Answer: 6.34
Example 3 · level 3
A planet has 4 times Earth's mass and 2 times Earth's radius. If Earth's escape speed is 11.2 km/s, find the planet's escape speed (in km/s, correct to 2 decimal places).
Show solution
- ve′ = 11.2 × √(4/2) = 11.2 × 1.4142 = 15.84 km/s.
Answer: 15.84