Check out some similar questions!

human09fishie · Oct 16, 2008, 03:19 PM

What is the distance from the Earth's center to a point outside the Earth where the gravitational acceleration due to the Earth is 1/14 of its value at the Earth's surface?

ebaines · Oct 17, 2008, 07:41 AM

Remmeber that gravity forces are inversely proportional to the distance to the object's center of gravity squared. That is:

$<br /> F \prop 1/r^2<br />$

If you call the radius of the earth R, then the gravity at the earth's surface (where r=R) is 1/R^2. So you're interested in finding the value of r such that :

$<br /> \frac 1 {r^2} = \frac 1 {14} \cdot \frac 1 {R^2}<br />$

Can you take it from here?

Credendovidis · Oct 18, 2008, 04:48 AM

Hello human09fishie

Have a look here for all data and explanation needed to calculate the distance - LINK

Success !

:)

.

.