91Ó°ÊÓ

The given data can be listed below,

• The radius of Deimos is, $r=\frac{12km}{2}\left(\frac{1000m}{1km}\right)$.
• The mass of Deimos is, $M=1.5x{10}^{15}\mathrm{kg}$.

When the gravitational attraction balances a satellite's speed, it circles that planet. Without this equilibrium, the satellite would either fly out into space or crash back to the planet.

From the law of conservation of energy,

$\frac{GMm}{r^2}=\frac{m{v}^2}{r}$

Here,Mis the mass of the Deimos, G is the constant of gravitation whose value is $6.673×{10}^{-11}N·m^2/k{g}^2$,m is the mass of the baseball, r is the radius of the Deimos, and v is the orbital speed of the baseball.

By solving the above equation,

role="math" localid="1668064719811" $v=\sqrt{\frac{GM}{r}}$

Substitute all values in the above,

$\begin{array}{r}v=\sqrt{\frac{\left(6.673×{10}^{−11}\mathrm{N}â‹\dots {\mathrm{m}}^2/{\mathrm{kg}}^2\right)\left(1.5×{10}^{15}\mathrm{kg}\right)}{6×{10}^3\mathrm{m}}\left(\frac{1\mathrm{kg}â‹\dots \mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}}\right)}\\ =4.07\mathrm{m}/\mathrm{s}\end{array}$

Thus, the speed required by baseball to achieve circular orbit and return is.

The ball can easily be thrown at this speed.

The time period of the orbit is given by,

$T=\frac{2\mathrm{Ï€°ù}}{v}$

Here, r is the radius of the Deimos, and v is the orbital speed of the baseball.

Substitute all the values in the above,

$\begin{array}{r}T=\frac{2\mathrm{π}\left(6×{10}^3\mathrm{m}\right)}{4.07\mathrm{m}/\mathrm{s}}\\ =9263\mathrm{s}\left(\frac{1\mathrm{hr}}{3600\mathrm{s}}\right)\\ =2.6\mathrm{hr}\end{array}$

Thus, the time taken by the baseball to return and hit is 2.6 hr. Yes, the game will last for a long time because the time of returning is significant.