## Essential University Physics: Volume 1 (3rd Edition)

We find the total energy: $E =- \frac{GMm}{2r}= - \frac{GM(40\times10^6kg)}{2(6.37\times10^6+35,000,000)}= -1.92\times10^{14}$ For the second one, we find: $E =- \frac{GMm}{2r}= - \frac{GM(12\times10^6kg)}{2(6.37\times10^6+30,000,000)}= -1.92\times10^{14}=-6.57\times10^{13}$ The fact that both values are negative tells us they are in solar orbit, meaning they will not strike earth.