Energy required to move an object to Earth's surface

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Homework Help Overview

The discussion revolves around calculating the energy required to move an object to Earth's surface, focusing on gravitational potential energy and the context of the object's initial position, potentially inside the Earth.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore different formulations of gravitational potential energy, questioning the initial conditions and the interpretation of the answer key. There are attempts to derive energy values based on varying assumptions about the object's starting position.

Discussion Status

Participants are actively engaging with the problem, offering calculations and questioning the correctness of the provided answer key. Some have provided alternative calculations based on different assumptions, while others express uncertainty about the initial conditions and the implications for the energy calculations.

Contextual Notes

There is a noted discrepancy between the calculated values and the answer key, with participants suggesting that the answer key may be incorrect. The discussion includes references to units of measurement, indicating a focus on clarity in the context of the problem.

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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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