Respuesta :
Answer: The height of the satellite above the surface of Earth is [tex]8.02\times 10^{5}m[/tex]
Explanation:
Given
Mass of the satellite, m= 1160 kg
tangential speed , v = 7446 m/s
Centripetal force , F = 8955 N
Radius of earth , R= [tex]6.38\times 10^{6}[/tex] m
Let height of satellite above surface of the Earth be H
Centripetal force on satellite is given by
[tex]F=\frac{mv^{2}}{R+H}[/tex]
=>[tex]H=\frac{mv^{2}}{F}-R[/tex]
=>[tex]H=(\frac{1160\times 7446^{2}}{8955}-6.38\times 10^{6}) m=8.02\times 10^{5}m[/tex]
Thus the height of the satellite above the surface of Earth is [tex]8.02\times 10^{5}m[/tex]
Answer:
[tex]h = 8.14 \times 10^5 m[/tex]
Explanation:
Let the height above the surface of Earth is H
now we know that gravitational force due to Earth on the satellite is net centripetal force due to which it is revolving around the Earth
So we can say
[tex]F = \frac{mv^2}{R}[/tex]
[tex]\frac{GM m}{(R + h)^2} = \frac{mv^2}{(R + h)}[/tex]
now we know that
[tex]\frac{GM}{(R + h)} = v^2[/tex]
[tex]\frac{(6.67 \times 10^{-11})(5.98 \times 10^{24})}{(6.38 \times 10^6 + h)} = 7446^2[/tex]
[tex]h = 8.14 \times 10^5 m[/tex]
