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?

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


F \prop 1/r^2


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 :


\frac 1 {r^2} = \frac 1 {14} \cdot \frac 1 {R^2}


Can you take it from here?

Hello human09fishie

Have a look here for all data and explanation needed to calculate the distance - LINK (http://www.glenbrook.k12.il.us/GBSSCI/PHYS/CLASS/circles/u6l3c.html)

Success !

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