Calculating Gravity at Half Earth's Radius

[bon]

Homework Statement

Taking the internal density of the Earth to be uniform, calculate, in terms of
the Earth’s radius R, the distance external to the surface of the Earth at which the
acceleration due to gravity is equal to that within the Earth at half the Earth’s radius.

Homework Equations

The Attempt at a Solution

Not sure about this one..Inside the Earth surely the bits of mass further out will pull out on an object.. so how do i work this out?

Thanks!

 

[Lok]

Hint! It is assumed, and mostly true that inside a hollow sphere of mass the forces acting upon an object cancel each other, wherever it is placed. So something like zero gravity.

 

[bon]

but the Earth isn't a hollow sphere :(

 

[Antiphon]

No, but if you hollowed out a small cave and made your measurement in there, you'd be on the right track. Additional hint: the "force of gravity" is zero in the center of the earth.

 

[Lok]

but the Earth isn't a hollow sphere :(

yes, but if you are at a certain distance inside the Earth then all that mass above you would not contribute to any gravity felt by you, only the mass below you. ( the sphere with radius= Earth radius-depth, while the hollow shell of depth thickness would not contribute)

 

[bon]

yes, but if you are at a certain distance inside the Earth then all that mass above you would not contribute to any gravity felt by you, only the mass below you. ( the sphere with radius= Earth radius-depth, while the hollow shell of depth thickness would not contribute)

why would the mass above you not contribute sorry? or is this just a theorem i should accept? :P

 

[americanforest]

why would the mass above you not contribute sorry? or is this just a theorem i should accept? :P

Gauss's Law for gravity. See proofs here if you are interested: http://en.wikipedia.org/wiki/Gauss's_law_for_gravity