Energy required to move an object to Earth's surface

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The discussion centers on calculating the energy required to move an object from inside the Earth to its surface. The initial calculations led to an incorrect result of -20R, while the answer key suggested 40R, indicating a misunderstanding of the starting position. After reevaluating the potential energy inside the Earth, the correct energy required was determined to be 7.5R. Participants noted that the answer should specify units in Newtons, highlighting a potential error in the provided answer. The conclusion reached is that the original answer key is incorrect.
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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
 
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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