Assuming only radial variations in planetary density, it can be shown that if a planet has a radius R, then to an observer at a distance r(> R) from the centre of the planet, all mass within the planet acts as if it is concentrated at the planet's centre. The escape velocity of an object placed in the gravitational field of a planet is given by Vescape = radical p2GM/r, where M is the total mass of the planet and G is the gravitational constant. Calculate the escape velocity at the Earth's surface. How does this compare to the escape velocity 100 km above the Earth's surface? How does that last figure compare to the velocity of a satellite in a circular orbit 100 km above the Earth's surface?
2007-10-04 14:44:50 · 1 answers · asked by -- 1 in Science & Mathematics ➔ Astronomy & Space
1) Use mass = 5.9742x10^24 kg,
Radius (mean) = 6371 km = 6,371,000 m
G = 6.672x10^-11 N m^2 / kg^2
V_e = SQRT( 2*G*M / r)
2) Long way = use r = 6371 + 100 km.
Short way: use answer from 1, multiply by SQRT(6371/6471) (escape speed varies as the square root of the inverse of the distance from the centre).
3) escape/orbital = √2
Use V_o = SQRT(G*M/r)
This is an approximation. It is OK if the mass of the satellite is negligible compared to that of the planet.
2007-10-04 14:59:15 · answer #1 · answered by Raymond 7 · 1⤊ 0⤋