# On earth, two parts of a space probe weigh 12500 N and 8400 N. These parts are separated by a center-to-center distance of 23 m and are spherical. How do you find the magnitude of the gravitational force that each part exerts on the other out in space?

Nov 2, 2015

$0.13$ microNewtons
And that's the same force they exert on each other on earth when set 23 meters apart.

#### Explanation:

Let's assume that $g = 10 \frac{m}{s} ^ 2$ to simplify some of our math.

The gravitational force between two objects can be calculates as a function of the product of their masses ($M$ and $m$), the square of the distance between them ($r$), and the universal gravitational constant. In the case of spherical objects, this is exactly correct. For more complicated shapes you might have to analyze different parts separately.

$F = G \frac{M m}{r} ^ 2$

The mass of the objects can be found by dividing their weight by the gravitational acceleration at the surface of the earth. I'm using $10 \frac{m}{s} ^ 2$ to make the mass easy.

$M = \frac{12500 N}{10 \frac{m}{s} ^ 2} = 1250 k g$
$m = \frac{8400 N}{10 \frac{m}{s} ^ 2} = 840 k g$

The distance was given:
$r = 23 m$

We can look up a value for G:
G = 6.67408 × 10^-11 m^3/ (kg s^2)

And plug that all into the first equation:
F = 6.67408 × 10^-11 (1250*840)/23^2 N
F = 1.325 × 10^-7 N

That seems very small, but in space, that could eventually draw these two objects together.