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The radius of the Earth can be estimated from its mass by using the acceleration due to the gravitational force of attraction that a body of mass 1 kg placed at sea level experiences. This is equal to 9.8 m/s^2.
For two bodies with mass M1 and M2 that are separated by a distance r, the force of attraction between them due to gravity is equal to F = G*M1*M2/r^2. Here G is a constant equal to 6.674*10^-11 N*m^2/kg^2. It has to be kept in mind that r is the distance between the center of gravity of each of the bodies.
Let the radius of the Earth be equal to R. A body at sea level is at a distance R from the center of gravity of the Earth. If the body weighs 1 kg, the gravitational force between the body and the Earth is equal to:
F = 6.674*10^-11*6*10^24*1/R^2
This force accelerates the body by 9.8 m/s^2. This also gives the force of attraction equal to 9.8 N
=> 6.674*10^-11*6*10^24*1/R^2 = 9.8
=> R^2 = 6.674*10^-11*6*10^24*1/9.8
=> R^2 = 4.086*10^13
=> R = 6.39*10^6 m
The radius of the Earth is approximately 6.39*10^6 m
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