Answer to Question #259613 in Mechanics | Relativity for Jay ar

Question

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.

Answer 1

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

$$\Delta PE=PE_f-PE_i,$$ $$\Delta PE=-\dfrac{GM_Em}{r_f}-(-\dfrac{GM_Em}{r_i}),$$ $$\Delta PE=GM_Em(-\dfrac{1}{r_f}+\dfrac{1}{r_i}).$$

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

$$\Delta PE=-\dfrac{GM_Em}{r_f},$$ $$\Delta PE=-\dfrac{6.67\times10^{-11}\ \dfrac{N\times m^2}{kg^2}\times5.98\times10^{24}\ kg\times1.0\times10^9\ kg}{4.0\times10^8\ m},$$ $$\Delta PE=-9.97\times10^{14}\ J.$$

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

$$W_{grav}=-\Delta PE=-(-9.97\times10^{14}\ J)=9.97\times10^{14}\ J.$$