Having aced your introductory physics course, you are hired as a summer intern a

Question

Having aced your introductory physics course, you are hired as a summer intern at NASA. You are sent as part of a team to explore the possibility of mining iron from a small, airless spherical asteroid with a radius of 458 km and a surface acceleration of 2.9 m/sec^2. 1) What is the mass of the asteroid? 2) What is the minimum vertical velocity you could give a 3 kg rock from the surface of this asteroid so that it continues to move upwards forever (i.e. it would never return to the surface)? 3) What is the minimum vertical velocity you could give a 6 kg rock from the surface of this asteroid so that it continues to move upwards forever (i.e. it would never return to the surface)? 4) How far will the 3 kg rock go if it leaves the asteroids surface with a vertical velocity of 1000 m/sec? 5) What will the speed of the 6 kg rock be at the surface of the asteroid if it is dropped from a point 1000km above the surface of the asteroid?

Explanation / Answer

Part 1)

Apply g = GM/r^2

(2.9)= (6.67 X 10^-11)(M)/(458000)^2

M = 9.12 X 10^21 kg

Part 2)

Escape Speed = sqrt(2GM/r)

v = sqrt[(2)(6.67 X 10^-11)(9.12 X 10^21)/(458000)]

v = 1.63 X 10^3 m/s

Part 3)

The escape speed is independent of mass, so its that same answer as part 2.

Part 4)

Conservation of Energy

GMm/r = .5mv^2

(6.67 X 10^-11)(9.12 X 10^21)/(r) = (.5)(1000)^2

r = 1.22 X 10^6 m

That distance is from the center of the asteroid. If we want the altitude from it surface, subract the radius of the asteroid

1.22 X 10^6 - 458000 = 7.59 X 10^5 m

Part 5)

GMm/r = .5mv^2

(6.67 X 10^-11)(9.12 X 10^21)/(1.46 X 10^6) = .5v^2

v = 913 m/s

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