An artificial satellite circles around the earth at a height where the gravitational force is th of that at the surface of the earth. if the earth's radius is r, the height of the satellite above the surface of the earth is :-
Answers
The escape velocity from earth is given by
v_{e} = \sqrt {2gR_{e}} ..... (i)v
e
=
2gR
e
.....(i)
The orbital velocity of a satellite revolving around earth is given by
v_{0} = \dfrac {\sqrt {GM_{e}}}{(R_{e} + h)}v
0
=
(R
e
+h)
GM
e
where, M_{e} =M
e
= mass of earth, R_{e} =R
e
= radius of earth, h =h= height of satellite from surface of earth.
By the relation GM_{e} = gR_{e}^{2}GM
e
=gR
e
2
So, v_{0} = \dfrac {\sqrt {gR_{e}^{2}}}{(R_{e} + h)} ..... (ii)v
0
=
(R
e
+h)
gR
e
2
.....(ii)
Dividing equation (i) by (ii), we get
\dfrac {v_{e}}{v_{0}} = \dfrac {\sqrt {2(R_{e} + h)}}{(R_{e})}
v
0
v
e
=
(R
e
)
2(R
e
+h)
Given, v_{0} = \dfrac {v_{e}}{2}v
0
=
2
v
e
\dfrac {2v_{e}}{v_{e}} = \dfrac {\sqrt {2(R_{e} + h)}}{R_{e}}
v
e
2v
e
=
R
e
2(R
e
+h)
Squaring on both side, we get
4 = \dfrac {2(R_{e} + h)}{R_{e}}4=
R
e
2(R
e
+h)
or R_{e} + h = 2R_{e}R
e
+h=2R
e
.
\implies\ h =R_e⟹ h=R
e