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AP EAMCET Engineering 22 Apr 2019 Shift 1 Paper
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© examsnet.com
Question : 92
Total: 160
A rocket is launched straight up from the surface of the earth. When its altitude is
1
3
of the radius of the earth, its fuel runs out and therefore it coasts. If the rocket has to escape from the gravitational pull of the earth, the minimum velocity with which it should coast is ( Escape velocity on the surface of the earth is 11.2
k
m
s
−
1
)
11.2
k
m
s
−
1
10.7
k
m
s
−
1
9.7
k
m
s
−
1
8.7
k
m
s
−
1
Validate
Solution:
Given the escape velocity on the surface of earth is,
v
e
=
11.2
km
∕
s
or
,
√
2
g
R
e
=
11.2
Also,
h
=
R
e
3
For the equilibrium, the potential energy at the altitude
R
e
3
should be equal to the kinetic energy.
Thus,
G
M
e
m
R
e
+
h
=
1
2
m
v
e
1
2
G
M
e
m
R
e
+
R
e
3
=
1
2
m
v
e
1
2
g
R
e
2
4
R
e
3
=
1
2
v
e
1
2
(
As
,
g
=
G
M
e
R
e
2
)
3
4
g
R
e
=
v
e
1
2
2
Solve further,
3
4
(
2
g
R
e
)
=
v
e
1
2
√
3
2
√
2
g
R
e
=
v
e
1
√
3
2
(
11.2
)
=
v
e
1
v
e
1
=
9.7
km
∕
s
© examsnet.com
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