An asteroid is discovered to have a tiny moon that orbits it in a circular path at a distance of \(100 . \mathrm{km}\) and with a period of \(40.0 \mathrm{~h}\). The asteroid is roughly spherical (unusual for such a small body) with a radius of \(20.0 \mathrm{~km} .\) a) Find the acceleration of gravity at the surface of the asteroid. b) Find the escape velocity from the asteroid.

We can use the given information about the moon's orbit to help us calculate the mass of the asteroid. According to Kepler's Third Law of Planetary Motion, the square of the period of the orbit is proportional to the cube of the semi-major axis. The formula for this relationship is: $$T^2 = \frac{4\pi^2}{G(M+m)}a^3$$ where \(T\) is the orbital period, \(a\) is the distance between the celestial bodies, \(G\) is the gravitational constant, and \(M\) and \(m\) are the masses of the two celestial bodies (in this case, the asteroid and its moon). In our case, \(m\) is negligible compared to \(M\), so we can approximate the equation as: $$T^2 = \frac{4\pi^2}{GM}a^3$$ We are given the orbital period \(T = 40.0 \mathrm{~h}=144000\mathrm{~s}\) and distance \(a = 100 \mathrm{~km}=1\times10^{5}\mathrm{~m}\). We can use these values to calculate the mass \(M\) of the asteroid. Rearrange the equation to solve for \(M\): $$M = \frac{4\pi^2a^3}{GT^2}$$

Now we plug in the values to calculate the mass of the asteroid: $$M = \frac{4\pi^2(1\times10^{5}\mathrm{~m})^3}{(6.674\times10^{-11}\mathrm{m^3~kg^{-1}~s^{-2}})(144000\mathrm{~s})^2}$$ $$M \approx 1.11\times10^{12}\mathrm{~kg}$$

Now that we have the mass of the asteroid, we can calculate the acceleration of gravity at the surface using the formula: $$g=\frac{GM}{r^2}$$ where \(r\) is the radius of the asteroid (\(20.0\mathrm{~km}=2\times10^{4}\mathrm{~m}\)). Plugging in the values: $$g=\frac{(6.674\times10^{-11}\mathrm{m^3~kg^{-1}~s^{-2}})(1.11\times10^{12}\mathrm{~kg})}{(2\times10^{4}\mathrm{~m})^2}$$ $$g\approx 1.85\times10^{-3}\mathrm{m ~s^{-2}}$$ So, the acceleration of gravity at the surface of the asteroid is approximately \(1.85\times10^{-3}\mathrm{m ~s^{-2}}\).

Finally, we calculate the escape velocity using the formula: $$v_{esc}=\sqrt{\frac{2GM}{r}}$$ Plugging in the known values: $$v_{esc}=\sqrt{\frac{2(6.674\times10^{-11}\mathrm{m^3~kg^{-1}~s^{-2}})(1.11\times10^{12}\mathrm{~kg})}{2\times10^{4}\mathrm{~m}}}$$ $$v_{esc}\approx 107\mathrm{~m ~s^{-1}}$$ The escape velocity from the asteroid is approximately \(107\mathrm{~m ~s^{-1}}\).