Unit 3 - Problem 8 => Sky diving & the earth

Given:

• 10 minute sky dive
• Mass of sky diver is 65 kg
• Mass of the earth $6\times10^{24}$

Reasoning:

• Since opposing forces must always be equal (for bodies at rest or constant velocity), then the force on the sky diver must also be acting on the earth.
• Note the time is 10 minutes, there for or calculations must use $(10\;minutes)\times60\;seconds = 600 seconds$.

Calculations:

$$F_{skydiver}=F_{earth}$$
$$F_{skydiver}=m_{skydiver}*a_{skydiver}=65\;{kg}*10\;{\frac{m}{s^2}}=650\;N$$
$$F_{earth}={6*10^{24}}*a_{earth}=650\;N$$
$$a_{earth}=\frac{650}{6*10^{24}}$$
$$x_{earth}=\frac{1}{2}*a_{earth}*t^2$$
$$x_{earth}=(.5*\left(\frac{650}{6*10^{24}}\right)*{600}^2$$
$$x_{earth}=195*10^{-19}=1.95*10^{-17}$$