If a 50 KG object is at a location 25,600 km from Earth's Center, what is the gravitational force exerted by the objects on Earth? In what direction does that force act? Support your answer with evidence.

The force exerted by the object on Earth is 30.4 N, and the direction is towards the object.

Explanation:

The gravitational force between two objects of mass M1 and M2, that are at a distance R between them is written as:

F = G*(M1*M2)/R^2

Where G is the gravitational constant, such that:

G = 6.674*1^(-11) m^3/(kg*s^2)

We know that M1, the mass of the object, is 50kg

R is the distance, in this case, is 25,600km

But this needs to be written in meters, remembering that:

1km = 1000m

Then:

25,600km = (25,600*1000)m = 25,600,000 m

And M2 is the mass of Earth, which is:

M2 = 5.972*10^(24) kg

Replacing all of those in the force equation we get

F = (6.674*1^(-11) m^3/(kg*s^2))*(50kg*5.972*10^(24) kg)/( 25,600,000 m)^2

F = 30.4 N

We know that the gravitational force is attractive, then the direction in which this force acts is towards the 50kg object.

Now, remember the second Newton's law is:

F = M*a

Then the acceleration that the object causes on Earth is:

30.4N = ( 5.972*10^(24) kg)*a

a = 30.4N/( ( 5.972*10^(24) kg)) = 5.09*10^(-24) m/s

This acceleration is almost despreciable.

the moon affects the earth gravitation ally, causing certain motions of that body and also producing tides in its ocean and atmosphere. the tidal wave in our atmosphere is, however, far too minute to be an effective cause in any known meteorological phenomenon.

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