Answer:
The magnitude of the free-fall acceleration at the orbit of the Moon is 
(
, where 
).
Explanation:
According to the Newton's Law of Gravitation, free fall acceleration (
), in meters per square second, is directly proportional to the mass of the Earth (
), in kilograms, and inversely proportional to the distance from the center of the Earth (
), in meters:
(1)
Where:
- Gravitational constant, in cubic meters per kilogram-square second.
- Mass of the Earth, in kilograms.
- Distance from the center of the Earth, in meters.
If we know that 
, 
and 
, then the free-fall acceleration at the orbit of the Moon is:
Answer:
a) 
, b) 
Explanation:
a) The final velocity of the 13.5 g coin is found by the Principle of Momentum Conservation:
The final velocity is:
b) The change in the kinetic energy of the 13.5 g coin is:
![\Delta K = \frac{1}{2}\cdot (13.5\times 10^{-3}\,kg)\cdot \left[(11.9\times 10^{-2}\,\frac{m}{s} )^{2}-(0\,\frac{m}{s} )^{2}\right]](https://tex.z-dn.net/?f=%5CDelta%20K%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20%2813.5%5Ctimes%2010%5E%7B-3%7D%5C%2Ckg%29%5Ccdot%20%5Cleft%5B%2811.9%5Ctimes%2010%5E%7B-2%7D%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%29%5E%7B2%7D-%280%5C%2C%5Cfrac%7Bm%7D%7Bs%7D%20%29%5E%7B2%7D%5Cright%5D)
Choice A is the right answer! Buy at high altitude and sell at low altitude
Hope this helps :)
Answer:
Haven't done this but I think it will increase by 73.29% not too anyone correct me if I'm worng
Well it seems that you did not give answer choices, but that its fine since we can use newtons law of universal gravitational, Fg = GM1M2/r^2. So G is the gravitational constant, which is 6.67*10^-11, we can plug in 6*1024 for M1, and 7*1022 for M2, and 3.8*108 for r. Which then we get 1.74 * 10^8 N as the force of attraction between the Earth and the moon.
