Question #259613

An asteroid woth mass = 1.00 x 109 kg comes from deep space, effectively from infinity, falls toward the Earth.




a. Find the change in potential energy when it reaches a point 4.00 x 108 m from the center of the earth.


b. Find the work done by the force of gravity.

1
Expert's answer
2021-11-01T12:53:24-0400

(a) From the definition of the gravitational potential energy, we have:


ΔPE=PEfPEi,\Delta PE=PE_f-PE_i,ΔPE=GMEmrf(GMEmri),\Delta PE=-\dfrac{GM_Em}{r_f}-(-\dfrac{GM_Em}{r_i}),ΔPE=GMEm(1rf+1ri).\Delta PE=GM_Em(-\dfrac{1}{r_f}+\dfrac{1}{r_i}).

Since the asteroid comes effectively from infinity, the term 1ri\dfrac{1}{r_i} is zero and we get:


ΔPE=GMEmrf,\Delta PE=-\dfrac{GM_Em}{r_f},ΔPE=6.67×1011 N×m2kg2×5.98×1024 kg×1.0×109 kg4.0×108 m,\Delta PE=-\dfrac{6.67\times10^{-11}\ \dfrac{N\times m^2}{kg^2}\times5.98\times10^{24}\ kg\times1.0\times10^9\ kg}{4.0\times10^8\ m},ΔPE=9.97×1014 J.\Delta PE=-9.97\times10^{14}\ J.

(b) By the definition of the work done by a conservative force, we have:


Wgrav=ΔPE=(9.97×1014 J)=9.97×1014 J.W_{grav}=-\Delta PE=-(-9.97\times10^{14}\ J)=9.97\times10^{14}\ J.

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