Magnitude of gravitational acceleration at distance r

[shyguy79]

Homework Statement

Find an expression for the ratio g(r)/g(RE) of the magnitude of the gravitational acceleration at a distance r from the centre of the Earth to that at the surface of the Earth, for r > RE, i.e. for points above the surface of the Earth.

Homework Equations

F(grav) = -Gm1m2 / r^2

The Attempt at a Solution

I was thinking about making r the subject of the equation and hence becoming R(E) and then substituting into the original equation but it doesn't seem to show anything.

I've been looking at this for a while but nothing is clicking any pointers would be very much appreciated as always.

 

[Spinnor]

At the surface of the Earth the force between a mass m and the Earth is

F = mg = ma but that is also equal to GmM/R^2 where R is the radius of the Earth. If the mass m is not near the surface of the Earth R will be different as will the acceleration.

Can you put it together?

 

[shyguy79]

I'm kinda confused.. but it makes sense! Was looking at Fgrav = mg at the surface of the Earth and yet above the surface of the Earth Fgrav = -Gm1m2 / r^2. I have substituted F=mg into F= -Gm1m2 / r^2 to become mg=-Gm1m2 / r^2 and then g =-Gm/r^2 but don't see how this helps

Doesn't this essentially say that the force of gravity on a body above the surface of the Earth is equal to the mass of the planet x gravitational constant x the inverse square of the radius? Or am I barking up the wrong tree?

Should I be looking at dividing F= -Gm1m2 / r^2 by F=mg?

 

[gneill]

I'm kinda confused.. but it makes sense! Was looking at Fgrav = mg at the surface of the Earth and yet above the surface of the Earth Fgrav = -Gm1m2 / r^2. I have substituted F=mg into F= -Gm1m2 / r^2 to become mg=-Gm1m2 / r^2 and then g =-Gm/r^2 but don't see how this helps

Doesn't this essentially say that the force of gravity on a body above the surface of the Earth is equal to the mass of the planet x gravitational constant x the inverse square of the radius? Or am I barking up the wrong tree?

Should I be looking at dividing F= -Gm1m2 / r^2 by F=mg?

You should lay aside the formula f = mg for this problem; it's only applicable for motions that occur very close to the Earth's surface, and is in fact an approximation derived from the general formula, g(r) = G*M/r² for r equal to the radius of the Earth ± small amounts. The particular value of g(r) obtained with r = RE is assigned to the constant 'g'.

Why not begin by writing the general formula for the acceleration twice, one for each of the radii: r and RE. Then form the ratio of the two expressions, cancelling common terms.

 

[shyguy79]

Ok, here's what I've done..

Taken g(r) = -GM/r^2 and g(Re) = -GM/Re^2 so putting and cancelling out G and M then g(r)/g(Re) = Re^2/r^2

Does this look plausible? I've put in a few figures and when both g(re) and g(r) = 6x10^6m then the ratio is 1... The shape of the graph is also an inverse exponential...

 

[gneill]

Ok, here's what I've done..

Taken g(r) = -GM/r^2 and g(Re) = -GM/Re^2 so putting and cancelling out G and M then g(r)/g(Re) = Re^2/r^2

Does this look plausible? I've put in a few figures and when both g(re) and g(r) = 6x10^6m then the ratio is 1... The shape of the graph is also an inverse exponential...

Yes, that's fine.

 

[shyguy79]

Thank you very much for your help!