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)
[tex]V_{orbital} = \sqrt{\frac{GM_E}{R}}[/tex]
Here,
M = Mass of Earth
R = Distance from center to the satellite
Replacing with our values we have,
[tex]V_{orbital} = \sqrt{\frac{(6.67*10^{-11})(5.972*10^{24})}{(6370*10^3)+(6370*10^3)}}[/tex]
[tex]V_{orbital} = 5591.62m/s[/tex]
[tex]V_{orbital} = 5.591*10^3m/s[/tex]
PART B) The period of satellite is given as,
[tex]T = 2\pi \sqrt{\frac{r^3}{Gm_E}}[/tex]
[tex]T = \frac{2\pi r}{V_{orbital}}[/tex]
[tex]T = \frac{2\pi (2*6370*10^3)}{5.591*10^3}[/tex]
[tex]T = 238.61min[/tex]
PART C) The gravitational force on the satellite is given by,
[tex]F = ma[/tex]
[tex]F = \frac{1}{4} mg[/tex]
[tex]F = \frac{270*9.8}{4}[/tex]
[tex]F = 661.5N[/tex]
(a) The orbital speed of the satellite is 5591.62 m/s.
(b) The period of revolution is 238.47 minutes.
(c) The gravitational force on satellite 661.5 N.
(a) The orbital velocity is given by;
[tex]v_{o}=\sqrt{\frac{GM}{r} }[/tex]
Here, 'M' is the mass of the earth and 'r' is the distance is from the centre of the earth to the satellite.
[tex]v_o = \sqrt{\frac{(6.67 \times 10^{-11})\times (5.972\times 10^{24})}{2\times 6370\times 10^{3}} } =5591.62\,m/s[/tex]
(b) The period of revolution is given by;
[tex]T=\frac{2\pi r}{V_o}=\frac{2\times 3.14\times 2\times 6370\times 10^3}{5591.62} =14308.411\,s = 238.47\,min[/tex]
(c) The gravitational force on the satellite is given by;
[tex]F_g = mg_s[/tex]
Where [tex]g_s[/tex] is the acceleration due to gravity at the satellite's height.
Given that, [tex]g_s = \frac{g}{4}[/tex]
[tex]F_g = \frac{270\times 9.8}{4} =661.5\,N[/tex]
Learn more about orbital motion here:
https://brainly.com/question/22247460
