When a falling meteoroid is at a distance above the earth's surface of 2.60 times the earth's radius, what is its acceleration due to the earth's gravitation?

The gravitational acceleration at any distance r is given by

where G is the gravitational constant, M the Earth's mass and r is the distance measured from the center of the Earth.

The Earth's radius is , so the meteoroid is located at a distance of:

And by substituting this value into the previous formula, we can find the value of g at that altitude:

g =0.76 m/s^2

Explanation: The gravitational acceleration is calculated with the equation.

g = G*M/r^

Where G is a constant of gravitation, M is the mass of the Earth and r is the distance of the meteoroid to the earth (to the center of the earth).

G = 6.67408×10^-11 m^3*kg^-1*s^-2

Earth radius = 6371km

M = 5.972 × 10^24 kg

and we know that r = 2.6 times the radius of the earth + radius of the earth, so r = 3.6*6371km

and we need to write this in meters, so r = 3.6*6371*1000m

and g = (6.67408×10^-11 m^3*kg^-1*s^-2)( 5.972 × 10^24 kg)/(3.6*1000*6371m)^2

= 0.76 m/s^2

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