Answer:
(a) T = 2.23h
(b) v = 6.78 km/s
(c) a = 5.31 m/s²
Explanation:
(a) The period (T) of the orbit can be calculated using Kepler's Third Law equation:
[tex] T = 2\pi \sqrt \frac{r^{3}}{GM} [/tex] (1)
where r: is the distance from the center of the Earth and the satellite, G: is the gravitational constant = 6.67x10⁻¹¹m³kg⁻¹s⁻² and M: is the Earth's mass = 5.97x10²⁴kg
Since [tex] r = r_{E} + r_{S} = 6.36\cdot 10^{6}m + 2.30\cdot 10^{6}m = 8.66\cdot 10^{6}m [/tex]
where [tex]r_{E}[/tex]: is the Earth's radius and [tex]r_{S}[/tex]: is the distance between the surface of the Earth and the satellite
Hence, by entering the radius calculated into equation (1) we can find the period of the orbit:
[tex] T = 2\pi \sqrt \frac{(8.66\cdot 10^{6}m)^{3}}{6.67\cdot 10^{-11}m^{3}/kgs^{2}\cdot 5.97\cdot 10^{24}kg} = 8024.3s = 2.23h [/tex]
(b) The speed (v) of the satellite can be calculated using the following equation:
[tex] v = \sqrt \frac{GM}{r} = \sqrt \frac{6.67\cdot 10^{-11}m^{3}/kgs^{2} \cdot 5.97\cdot 10^{24}kg}{8.66 \cdot 10^{6}m} = 6.78km/s [/tex]
(c) The acceleration (a) of the satellite is:
[tex] a = \frac{v^{2}}{r} = \frac{(6.78 \cdot 10^{3}m/s)^{2}}{8.66\cdot 10^{6}m} = 5.31m/s^{2} [/tex]
I hope it helps you!
