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Imagine that you are prospecting for rare metals on a spherical asteroid compose

ID: 2184792 • Letter: I

Question

Imagine that you are prospecting for rare metals on a spherical asteroid composed mostly of iron (density: 7800 kg/m^3) and whose radius is 4500 m. You've left your spaceship in a circular orbit 400 m above the asteroid's surface and gone down to the surface using a jetpack. However, one of your exploratory explosions knocks you back against a rock, ruining your jet pack. Is it possible for you to simply jump high enough in this situation to get back to the spaceship?

Can you please tell me how you get the answer. And your reasoning.

Explanation / Answer

1. Find the mass of the asteroid. 2. Use the mass of the earth, the radius of the earth, and the radius of the asteroid, to find the gravity at the asteroid surface, in units of ''g", where earth gravity = 1 g. Let: M = mass of asteroid, m = mass of earth, R = radius of asteroid, r = radius of earth: Asteroid gravity = g(M/R^2) / (m/r^2). 3. Decide how high you can jump on earth, and use this PE in Joules to compute the height you can attain on the asteroid. let's find the value of g, accel due to gravity on the asteroid we know that g=GM/R^2 where G=newtonian grav cst, M=mass of planet, R=radius of planet now, we don't even need to separately compute M... M=rho*volume = 4/3 pi R^3 rho so substitute this into the equation for g: g=G(4/3 pi rho R^3)/R^2 = 4/3 pi rho G R using values: g=4/3*3.14*6.67x10^(-11)*7800kg/m^3*45… g=0.01m/s/s now, what would your initial velocity have to be to reach a height of 400m? use conservation of energy 1/2 mv^2 = mgh (remember to use the local value of g) v=sqrt[2x0.01m/s/sx400m] v=2.8m/s (which is about 6.3 mi/hr...a speed most people can attain...so...yes!!...but don't miss your target) we have assumed that the value of g does not vary in 400m, this is probably ok since 400m is less than 1/10 the radius of the asteroid, so the value of g would not decrease by more than 1% (and in any rate, a smaller g makes it easier to reach the ship)