1.

An astronaut standing on the surface of the moon (m=M, radius=R) holds a feather (mass=m) in one hand and a hammer (mass=100m) in the other hand, both at the same height above the surface. If he releases them simultaneously, what is the acceleration of the hammer?

Answer»

`(mv^(2))/(R)`
`(GM)/(R^(2))`
`(GM m)/(R^(2))`
`100(GM)/(R^(2))`

Solution :The GRAVITATIONAL force on an object of mas m can be expressed either by MG or by `(GM m)/(r^(2))`. SETTING mg equal to `(GM m)/(r^(2))`, we get `g=(GM)/(R^(2))`, which is the object's (free-fall) aceleration. Notice that the mass of the object cancels out, so whether we're asked for the acceleration of the feather or the hammer, the answer would be the same.


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