Gravitation deep hole physics problem

[brad sue]

Hi I am stuck with this problem:
A deep hole in Earth reaches a depth of one half of Earth's radius. How much work is done when a 1-kg mass is slowly lifted from the bottom of the hole to Earth's surface?
I use the expression of the gravitation force, F=GMm/R3
so the work would be:
integral (F*(R/2)) from R=R/2 to R=R
I am not sure I am right so I need your input please.
Thank you
Brad

 

[ahrkron]

Make sure you include the dependence that M has on r (the integration variable).

 

[lightgrav]

Your Force function is incorrect.

The Gravitational Force has an INVERSE SQUARE form $$F = m \frac{GM}{r^2}$$ .
The Force by Gravity will only be due to the mass of Earth that is
INTERIOR to the mass being lifted - presume uniform mass density, that is $$M = M_E \frac{V_{inside}}{V_E}$$ .

 

[brad sue]

Your Force function is incorrect.

The Gravitational Force has an INVERSE SQUARE form $$F = m \frac{GM}{r^2}$$ .
The Force by Gravity will only be due to the mass of Earth that is
INTERIOR to the mass being lifted - presume uniform mass density, that is $$M = M_E \frac{V_{inside}}{V_E}$$ .

Thank you.
But How can I compute the V_inside? we only have the length nothing else?
B

 

[lightgrav]

Volume? LxWxH !
Yeah, rxrxr = r^3 , the variable r ,
compared with $$R_E^3$$ , the maximum R .

 

[brad sue]

Volume? LxWxH !
Yeah, rxrxr = r^3 , the variable r ,
compared with $$R_E^3$$ , the maximum R .

OK! I am sorry . sometimes I am a little bit way off!

thank you so much
I can finish this problem now.

B