Force = mass times acceleration
F = 21000 x 36.9 = 774900
Therefore, 774900N force is required.
All electromagnetic waves travel at the same speed in a vacuum: 3.0 x 10^5 (300,000) kilometres per second. some electromagnetic waves are part of the visible light spectrum and some do emit harmful radiation, but certainly not all. they travel fine on earth without the vacuum of space too.
To solve this problem we will apply the concepts related to Orbital Speed as a function of the universal gravitational constant, the mass of the planet and the orbital distance of the satellite. From finding the velocity it will be possible to calculate the period of the body and finally the gravitational force acting on the satellite.
PART A)
![V_{orbital} = \sqrt{\frac{GM_E}{R}}](https://tex.z-dn.net/?f=V_%7Borbital%7D%20%3D%20%5Csqrt%7B%5Cfrac%7BGM_E%7D%7BR%7D%7D)
Here,
M = Mass of Earth
R = Distance from center to the satellite
Replacing with our values we have,
![V_{orbital} = \sqrt{\frac{(6.67*10^{-11})(5.972*10^{24})}{(6370*10^3)+(6370*10^3)}}](https://tex.z-dn.net/?f=V_%7Borbital%7D%20%3D%20%5Csqrt%7B%5Cfrac%7B%286.67%2A10%5E%7B-11%7D%29%285.972%2A10%5E%7B24%7D%29%7D%7B%286370%2A10%5E3%29%2B%286370%2A10%5E3%29%7D%7D)
![V_{orbital} = 5591.62m/s](https://tex.z-dn.net/?f=V_%7Borbital%7D%20%3D%205591.62m%2Fs)
![V_{orbital} = 5.591*10^3m/s](https://tex.z-dn.net/?f=V_%7Borbital%7D%20%3D%205.591%2A10%5E3m%2Fs)
PART B) The period of satellite is given as,
![T = 2\pi \sqrt{\frac{r^3}{Gm_E}}](https://tex.z-dn.net/?f=T%20%3D%202%5Cpi%20%5Csqrt%7B%5Cfrac%7Br%5E3%7D%7BGm_E%7D%7D)
![T = \frac{2\pi r}{V_{orbital}}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B2%5Cpi%20r%7D%7BV_%7Borbital%7D%7D)
![T = \frac{2\pi (2*6370*10^3)}{5.591*10^3}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B2%5Cpi%20%282%2A6370%2A10%5E3%29%7D%7B5.591%2A10%5E3%7D)
![T = 238.61min](https://tex.z-dn.net/?f=T%20%3D%20238.61min)
PART C) The gravitational force on the satellite is given by,
![F = ma](https://tex.z-dn.net/?f=F%20%3D%20ma)
![F = \frac{1}{4} mg](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20mg)
![F = \frac{270*9.8}{4}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7B270%2A9.8%7D%7B4%7D)
![F = 661.5N](https://tex.z-dn.net/?f=F%20%3D%20661.5N)
Answer:
Explanation:
Answer
The true fact is that C is what happens in outer space. Both rotations take 27.3 days.
A: The exact opposite is true. It does rotate about it's axis.
B: Again this is just plain false. Given the way we observe it, the moon must be rotating around the earth.
D. they don't. 27.3 hours and 24 hours are not the same.