An astronaut finds himself on a spherical asteroid whose radius is 500 km and whose mass is 1.2 x 1022 kg. a) What is the escape speed from the surface? b) How far from the surface will a particle go if it leaves the asteroid's surface with a radial speed of 1000 m/s? c) With what speed will an object hit the asteroid if it is dropped from 1000 km above the surface? d) What is the acceleration due to the asteroid's gravitational field at the surface? Recall that g = GM/R2

a) v = 1.79 10³ m / s, b) R_max = 1.33 10⁶ m, c) v = 1.26 10³ m / s, d) g = 3.2016 m / s²

Explanation:

a) to find the escape velocity let us use energy conservation

Initial. On the surface

Em₀ = K + U = ½ m v² - G m M / R

Final. Far from the asteroid

Emf = U = - G m M / R_max

Em₀ = Emf

½ m v² - G m M / R = - G mM / R_max

To escape the asteroid the distance must be infinite, so the last term is zero

v = √2GM / R

v = √ (2 6.67 10⁻¹¹ 1.2 10²²/500 10³)

v = √ 3.2 10⁶

v = 1.79 10³ m / s

b) how high can it go if it leaves with a speed of v = 1000 m / s

½ m v² - G m M / R = - G mM / R_max

1 / R_max = 1 / R - ½ v² / GM

1 / R_max = 1/500 10³ - ½ 1000² / (6.67 10⁻¹¹ 1.2 10²²)

1 / R_max = 2 10-6 - 1,249 10-6 = 0.751 10-6

R_max = 1.33 10⁶ m

c) The speed if the object falls from h = 1000 km above the superfine

The distance from the center of the asteroid is

R ' = R + h

R ' = 500 + 1000 = 1500 km

R ' = 1.5 106 m

½ m v² - G m M / R = - G mM / R '

½ v² = GM (1 / R - 1 / R ')

v = √ 2 GM (1 / R - 1 / R ')

v = √ (2 6.67 10⁻¹¹ 1.2 10²² (1/5 10⁵ - 1/1 10⁶)

v = √ 1.6 10¹² (1 10⁻⁶)

v = 1.26 10³ m / s

d) the gravity acceleration of the asteroid is

g = G M / R²

g = 6.67 10⁻¹¹ 1.2 10²² / (5 10⁵) ²

g = 3.2016 m / s²