Answer:
a) a geostationary satellite is that it is always at the same point with respect to the planet,
b) f = 2.7777 10⁻⁵ Hz
c) d) w = 1.745 10⁻⁴ rad / s
Explanation:
a) The definition of a geostationary satellite is that it is always at the same point with respect to the planet, that is, its period of revolutions is the same as the period of the planet
- T = 10 h (3600 s / 1h) = 3.6 104 s
b) the period the frequency are related
T = 1 / f
f = 1 / T
f = 1 / 3.6 104
f = 2.7777 10⁻⁵ Hz
c) the distance traveled by the satellite in 1 day
The distance traveled is equal to the length of the circumference
d = 2pi (R + r)
d = 2pi (69 911 103 + 120 106)
d = 1193.24 m
d) the angular velocity is the angle traveled between the time used.
.w = 2pi /t
w = 2pi / 3.6 10⁴
w = 1.745 10⁻⁴ rad / s
how fast is
v = w r
v = 1.75 10-4 (69.911 106 + 120 106)
v = 190017 m / s
Answer:
The Answer is D!
Explanation:
I checked it on Khan Academy.
Answer: 
Explanation:
According to the <u>Third Kepler’s Law</u> of Planetary motion:
(1)
Where;:
is the period of the satellite
is the Gravitational Constant and its value is 
is the mass of the Earth
is the semimajor axis of the orbit the satllite describes around the Earth (as we know it is a circular orbit, the semimajor axis is equal to the radius of the orbit).
On the other hand, the orbital velocity 
is given by:
(2)
Now, from (1) we can find
, in order to substitute this value in (2):
![a=\sqrt[3]{\frac{T^{2}GM}{4\pi}^{2}}](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7B%5Cfrac%7BT%5E%7B2%7DGM%7D%7B4%5Cpi%7D%5E%7B2%7D%7D)
(3)
![a=\sqrt[3]{\frac{(1.26(10)^{4}s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{4\pi}^{2}}](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%281.26%2810%29%5E%7B4%7Ds%29%5E%7B2%7D%286.674%2810%29%5E%7B-11%7D%5Cfrac%7Bm%5E%7B3%7D%7D%7Bkgs%5E%7B2%7D%7D%29%285.98%2810%29%5E%7B24%7Dkg%29%7D%7B4%5Cpi%7D%5E%7B2%7D%7D)
(4)
(5)
Substituting (5) in (2):
(6)
(7) This is the speed at which the satellite travels
