What is the minimum energy required to launch a satellite of mass m from the surface of a planet of mass M and radius R in a circular orbit at an altitude of 2R?
Answers
Explanation:
The kinetic energy at altitude 2R=GMm6R
The gravitational potential energy at altitude
2R=−GMm3R
Total energy =k∈+PE=−GMm6R
Potential energy at the surface is −GMmR
∴ Req. kinetic energy =GMmR−GMm6R
= 5GMm6R
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ANSWER~
Answer
Given that,
Mass of satellite =m=m
Mass of planet =M=M
Radius =R=R
Altitude h=2Rh=2R
Now,
The gravitational potential energy
P.E=\dfrac{-Gm}{r}P.E=
r
−Gm
Potential energy at altitude =\dfrac{GmM}{3R}=
3R
GmM
Orbital velocity {{v}_{0}}=\dfrac{GmM}{R+h}v
0
=
R+h
GmM
Now, the total energy is
{{E}_{f}}=\dfrac{1}{2}mv_{0}^{2}-\dfrac{GmM}{3R} E
f
=
2
1
mv
0
2
−
3R
GmM
{{E}_{f}}=\dfrac{1}{2}\dfrac{GmM}{3R}-\dfrac{GMm}{3R} E
f
=
2
1
3R
GmM
−
3R
GMm
{{E}_{f}}=\dfrac{GmM}{3R}\left[ \dfrac{1}{2}-1 \right] E
f
=
3R
GmM
[
2
1
−1]
{{E}_{f}}=\dfrac{-GmM}{6R} E
f
=
6R
−GmM
Now, {{E}_{i}}={{E}_{f}}E
i
=E
f
Now, the minimum required energy
K.E=\dfrac{Gmm}{R}-\dfrac{GmM}{6R} K.E=
R
Gmm
−
6R
GmM
K.E=\dfrac{5GmM}{6R} K.E=
6R
5GmM
Hence, the minimum required energy is \dfrac{5GmM}{6R}
6R
5GmM