Calculating gravitational pull on surface for large object

[5P@N]

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?

 

[gneill]

Looks fine.