Question 8:The energy required to take a satellite to aheight 'h' above Earth surface (radius ofEarth =6.4 x 10 km is E, and kineticenergy required for the satellite to be in acircular orbit at this height is E. The valueofh for which E, and E, are equal, is.
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AnswerU
surface
+E
1
=U
h
KE of satelite is zero at earth surface & at height h
−
R
e
GM
e
m
+E
1
=−
(Re+h)
GM
e
m
E
1
=GM
e
m(
R
e
1
−
R
e
+h
1
)
E
1
=
(R
e
+h)
GM
e
m
×
R
e
h
Gravitational attraction F
G
=ma
C
=
(R
e
+h)
mv
2
E
2
⇒
(R
e
+h)
mv
2
=
(R
e
+h)
2
GM
e
m
mv
2
=
(R
e
+h)
GM
e
m
E
2
=
2
mv
2
=
2(R
e
+h)
GM
e
m
E
1
=E
2
R
e
h
=
2
1
⇒h=
2
R
e
=3200km
Explanation:pls mark it brainliest
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