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Newton's Law of Gravity
Climbing to the top of Mount Everest is hard, but it's slightly easier than you might think as people weigh less as they climb to the top. Let \(W_E\) be a person's weight on top of Mount Everest and \(W_S\) be their weight at sea level. What is the value of \(1-W_E/W_S\)?
Details and assumptions
- Assume the earth (other than Everest) is a sphere of mass \(6 \times 10^{24}~\mbox{kg}\) and radius \(6,370~\mbox{km}\).
- The top of Mount Everest is \(8,848~\mbox{m}\) above the surface of the earth.
Assume that the earth and the moon are both perfect spheres and the earth-moon distance is constant.
If the magnitude of gravitational attraction force between two particles with respective masses \(5.5\text{ kg}\) and \(2.9\text{ kg}\) is \(2.7 \times 10^{-12}\text{ N},\) what is the approximate separation between the two particles, given that the universal gravitational constant is \(G=6.67 \times 10^{-11} \text{ N}\cdot\text{m}^2\text{/kg}^2?\)
A neutron star has a mass equal to that of the sun, which is \(2.1 \times 10^{30}\text{ kg},\) but has a radius of only \(7\text{ km}.\) What is the approximate gravitational acceleration at the surface of the star?
Assumptions and details
- The universal gravitational constant is \(G=6.67 \times 10^{-11} \text{ N}\cdot\text{m}^2\text{/kg}^2.\)
- The universal gravitational constant is \(G=6.67 \times 10^{-11} \text{ N }\cdot\text{m}^2\text{/kg}^2.\)
- \(R = 5.5\times10^6\) m.