Assume this asteroid has the shape of a sphere. A rock sitting on the surface of the asteroid has a weight of 100 Newtons. What would the weight of this rock be if it were moved to a distance above...

Assume this asteroid has the shape of a sphere. A rock sitting on the surface of the asteroid has a weight of 100 Newtons. What would the weight of this rock be if it were moved to a distance above the surface equal to the radius of the asteroid? Include calculations or the logic used to get your answer. How is the mass of the rock affected by its change in position?

Asked on by jennie2412

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llltkl | College Teacher | (Level 3) Valedictorian

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Weight is the force with which the asteroid pulls rock towards its center. Hence, from Newton's law of gravitation:

`(GMm)/R^2=100` ..................(i)

Where G=universal gravitational constant , M=mass of the asteroid, m= mass of the rock and R= distance between the rock and the asteroid.

If distance becomes 2R  [it's 2 times the radius because it's moved a distance r and originally from the center of the astroid to
the surface the distance is r ] and W is the new weight of the rock, we have

`(GMm)/(2R)^2=W` .........................(ii)

Dividing (i) by (ii) and solving we get:

`1/4=100/W`

`rArr W=25`

Hence, the new weight of the rock is 25 N.

Mass of the rock doesn't get affected by its change in position. It remains constant everywhere in the universe, only its weight changes.

In general, if the distance between asteroid and rock is `(R+x)` and `W_x` is the weight, we have:

`(GMm)/(R+x)^2=W_x` ................(iii)

Diving (i) by (iii) and simplifying, we get:

`W_x=100[R/(R+x)]^2`

The above equation gives relation between weight and distance between asteroid and rock.

Sources:

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