Acceleration Part 2: Obtain Expression for g in Terms of G, Ra & ρ

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Homework Statement



density, ρ of the rock is 4000 kg m−3 and that this density is uniform throughout. You have equipment that enables you to measure the time it takes for a small object to fall from rest through a distance of 2.00 m. Your measurement is 8.0 s.

Obtain an expression for ga in terms of the gravitational constant G, the
radius of the asteroid Ra and the density ρ of the asteroid. Hence determine
Ra.

Homework Equations



F= Gm/ r2

F=mg

volume of sphere= 4/3 * pi * r^3

density= m/v

The Attempt at a Solution



substituting volume of sphere= 4/3 * pi * r^3 into density= m/v gives-

density= 3m/ ( 4* pi * r^3 )

4/3 * pi * r^3 * density = m

F= Gm/ r2

F= (G*4/3 * pi * r^3 * density) / r^2

F=G*4/3 * pi * r * density

F=mg

mg= G*4/3 * pi * r * density

4/3 * pi * r^3 * density = m

4/3 * pi * r^3 * density* g = G*4/3 * pi * r * density

g= Gr^-2

or g= G / r^2


(is this correct)?
 
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F= Gm/ r2 ?

No.
F = G*m1*m2/r^2
 
if i use F = G*m1*m2/r^2, i am only given 1 object, 1 mass...

so how would i go about doing this by using

F = G*m1*m2/r^2
 
You don't need to know the other mass, but you must include it in your equations. You'll see why.
 
imy786 said:

Homework Statement



density, ρ of the rock is 4000 kg m−3 and that this density is uniform throughout. You have equipment that enables you to measure the time it takes for a small object to fall from rest through a distance of 2.00 m. Your measurement is 8.0 s.

Obtain an expression for ga in terms of the gravitational constant G, the
radius of the asteroid Ra and the density ρ of the asteroid. Hence determine
Ra.

Homework Equations



F= Gm/ r2

F=mg

volume of sphere= 4/3 * pi * r^3

density= m/v

The Attempt at a Solution



substituting volume of sphere= 4/3 * pi * r^3 into density= m/v gives-

density= 3m/ ( 4* pi * r^3 )

4/3 * pi * r^3 * density = m

F= Gm/ r2

F= (G*4/3 * pi * r^3 * density) / r^2

F=G*4/3 * pi * r * density

F=mg

mg= G*4/3 * pi * r * density

4/3 * pi * r^3 * density = m

4/3 * pi * r^3 * density* g = G*4/3 * pi * r * density

g= Gr^-2

or g= G / r^2


(is this correct)?

As others pointed out, you are not using the correct equation for the force of gravity!

be very careful, there are two masses involved here, th etotal mass of the "rock and the mass of the falling object. You should really use two different symbols and be careful about using the correct one in each eqaution you are using.

As for your final answer, there is an obvious simple check to make...does it have the dimensions of an acceleration??
 
F = G*m1*m2/r^2

substituting volume of sphere= 4/3 * pi * r^3 into density= m/v gives-

density= 3m/ ( 4* pi * r^3 )

4/3 * pi * r^3 * density = m

im stuck here...

shall i definte 4/3 * pi * r^3 * density = m = m1

and leave m2 as m2
 
Sure. Just keep them consistent.
 
F = G*m1*m2/r^2

V= (4/3)*PI*r^3

p= m /v

p= m1/ (4/3)*PI*r^3

m1= p*4*PI*r^3/ 3

now we substiute this value of m1 to this equation F = G*m1*m2/r^2

F = G*m2*p*4*PI*r/ 3

is final answer...can someone tell me if this correct..to Obtain an expression for ga in terms of the gravitational constant G, the
radius of the asteroid Ra and the density ρ of the asteroid. Hence determine
Ra.
 
F = G*m2*p*4*PI*r/ 3

F=mg

mg= G*m2*p*4*PI*r/ 3

g= G*p*4*PI*r/ 3

as final answer...can anyone..confirm if this is correct formula

ga in terms of the gravitational constant G, the
radius of the asteroid Ra and the density ρ of the asteroid.
 
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