Question

Imagine that you are standing on a spherical asteroid deep in space far from other objects. You pick up a small rock and throw it straight up from the surface of the asteroid. The asteroid has a radius of 9 m and the rock you threw has a mass of 0.113 kg. You notice that if you throw the rock with a velocity less than 45.7 m/s it eventually comes crashing back into the asteroid.

Required:
Calculate the mass of the asteroid.

Answer 1

Answer:

M = 1.409 10¹⁴ kg

Explanation:

In this exercise we have that the prioress with a minimum speed can escape from the asteroid, therefore we can use the conservation of energy relation.

Starting point. When you drop the stone

         Em₀ = K + U

         Em₀ = ½ m v² - G m M / r

where M and r are the mass and radius of the asteroid

Final point. When the stone is too far from the asteroid

          Em_f = U = - G m M / R_f

as there is no friction, the energy is conserved

          Em₀ = Em_f

          ½ m v² - G m M / r = - G m M / R_f

          ½ v² = G M (1 / r - 1 /R_f)

indicate that for the speed of v = 45.7 m /s, the stone does not return to the asteroid so R_f = ∞

          ½ v² = G M (1 /r)

          M = $\frac{v^2 r}{2G}$

 

let's calculate

          M = $\frac{45.7^2 \ 9}{ 2 \ 6.67 \ 10^{-11}}$

          M = 1.409 10¹⁴ kg