## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

The mass of the planet is $4.08\times 10^{24}~kg$
We can find the free-fall acceleration $g'$ on this planet as: $y = \frac{1}{2}g't^2$ $g' = \frac{2y}{t^2}$ $g' = \frac{(2)(100~m)}{(6.0~s)^2}$ $g' = 5.56~m/s^2$ We then use the free-fall acceleration to find the mass $M_p$ of the planet. $\frac{G~M_p}{R^2} = g'$ $M_p = \frac{R^2~g'}{G}$ $M_p = \frac{(7.0\times 10^6~m)^2~(5.56~m/s^2)}{6.67\times 10^{-11}~m^3/kg~s^2}$ $M_p = 4.08\times 10^{24}~kg$ The mass of the planet is $4.08\times 10^{24}~kg$.