Gravity on Earth: 70% Value Above Surface?

[zzinfinity]

Homework Statement

How far above the surface of the Earth do you need to be for the force of gravity to be reduced to 70% of its value on the surface.

Homework Equations

g on surface is 9.8 m/s^2

Fg=G*m1*m2/r^2

The Attempt at a Solution

Is there a way to do this without knowing the mass of the earth? It seems as though the gravity equation I have above could only be useful if you knew either the mass of the Earth or the radius of the earth. I was just wondering if there is a way to approach this without knowing either of those. Thanks!

 

[SteamKing]

Instead of plug and chug into the gravitational force equation, try working with ratios.
Everything drops out except for the radii between the masses.

 

[Chi Meson]

Since you are being asked for the specific distance, then you will ultimately need to know the Earth's radius, which is easy enough to find.

Remember that "r" is not zero at Earth's surface, but "R" (let that represent the Earth's radius.

Think about the "inverse square proportionality" of force with distance. If distance doubled (that is, if you went a distance of R above the surface so the distance from Earth's center is "2R"), the field and therefor the force would be (1/4)F , or (0.25)F (compared to F on the surface). If distance was 3R, the force would be (1/9)F , or (0.111)F .

You need to figure out what factor of R would produce (0.70)F