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Homework Statement
A large sphere exists in space, which has a mass of 1 * 10^28 kg
The sphere has a radius of 100,000 km
What will be its gravitational pull (aka: "relative gravity") on its surface in terms of gs (1 "g" being equal to the gravitational pull of the Earth which is 9.807 m/s^2)?
Homework Equations
relative gravity = (gravitational constant * mass of object) / radius^2
gravitational constant = 6.67408 * 10^-11 m^3 kg^-1 s^-2
The Attempt at a Solution
First I multiplied the Gravitational Constant by the object's mass:
(6.67408 * 10^-11 m^3 kg^-1 s^-2) * (1*10^28 kg) = 6.67408 *10^17 m^3 s^-2
Next I squared the radius:
(100,000,000 m)^2 = 1*10^16 m^2
Then I divided the numerator by the denominator:
(6.67408 *10^17 m^3 s^-2) / (1*10^16 m^2) = 66.7408 m s^-2.
This I define as the relative gravity upon the sphere's surface.
Finally, I divided this relative gravity by the Earth's gravity:
(66.7408 m s^-2) / (9.807 m s^-2) = 6.805 "gs"
Thus: the gravitational pull upon this sphere's surface is about 6.805 times the gravity on the Earth's surface.
Is my methodology and/or reasoning correct ladies & gentlemen?