Energy required to move an object to Earth's surface

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SUMMARY

The energy required to move an object from inside the Earth to its surface is calculated using gravitational potential energy formulas. The correct expression for energy change is derived as ΔEp = 7.5R, indicating that the initial assumption of 40R was incorrect. The potential energy inside the Earth is given as -11/8 GMm/R, leading to the conclusion that the answer key provided was erroneous. The unit for the final answer should be specified as Newtons (N).

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Homework Statement
Distance of an object from center of earth is twice the radius of earth and the gravitational force acting on it is 10 N. Find the energy required to move the object to earth's surface
Relevant Equations
##F=\frac{GMm}{r^2}##

##E_p=-\frac{GMm}{r}##
Energy required = ##\Delta E_p##
$$\Delta E_p = -\frac{GMm}{R}+\frac{GMm}{2R}$$
$$=-\frac{1}{2} \frac{GMm}{R}$$
$$=-\frac{1}{2} \frac{GMm}{R^2} R$$
$$=-\frac{1}{2} 40R$$
$$=-20R$$

But the answer key is 40R. Where is my mistake?

Thanks
 
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I note the answer given is positive, suggesting it was supposed to be from a starting position inside the Earth. What do you get if you change "twice" to "half"?
 
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haruspex said:
I note the answer given is positive, suggesting it was supposed to be from a starting position inside the Earth. What do you get if you change "twice" to "half"?
The object will be inside Earth initially so:
$$F=\frac{GMm}{R^3}r$$
$$10=\frac{GMm}{R^3}\left(\frac{1}{2}R\right)$$
$$\frac{GMm}{R^2}=20$$

The potential inside Earth :
$$V=-\frac{GM(3R^2 - r^2)}{2R^3}$$
$$=-\frac{GM(3R^2 - 0.25R^2)}{2R^3}$$
$$=-\frac{11}{8} \frac{GM}{R}$$

The potential energy inside Earth is ##-\frac{11}{8} \frac{GMm}{R}##

Energy required:
$$\Delta E_p=-\frac{GMm}{R}+\frac{11}{8} \frac{GMm}{R}$$
$$=\frac{3}{8} \frac{GMm}{R}$$
$$=\frac{3}{8} \frac{GMm}{R^2}R$$
$$=7.5R$$

Is this correct?

Thanks
 
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songoku said:
The object will be inside Earth initially so:
$$F=\frac{GMm}{R^3}r$$
$$10=\frac{GMm}{R^3}\left(\frac{1}{2}R\right)$$
$$\frac{GMm}{R^2}=20$$

The potential inside Earth :
$$V=-\frac{GM(3R^2 - r^2)}{2R^3}$$
$$=-\frac{GM(3R^2 - 0.25R^2)}{2R^3}$$
$$=-\frac{11}{8} \frac{GM}{R}$$

The potential energy inside Earth is ##-\frac{11}{8} \frac{GMm}{R}##

Energy required:
$$\Delta E_p=-\frac{GMm}{R}+\frac{11}{8} \frac{GMm}{R}$$
$$=\frac{3}{8} \frac{GMm}{R}$$
$$=\frac{3}{8} \frac{GMm}{R^2}R$$
$$=7.5R$$

Is this correct?

Thanks
Yes, that's correct. Since it does not result in 40R N that doesn’t help. So it looks like the given answer is simply wrong.
Btw, the answer should specify the unit "Newtons", i.e. -20R N.
 
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Thank you very much haruspex
 
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