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BITSAT 2019 Slot 2 Solved Paper
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© examsnet.com
Question : 28
Total: 150
The escape velocity of the earth surface is
v
e
.
A body is projected with velocity
3
v
e
.
With what constant velocity will it move in the inter planetary space?
v
e
3
v
e
√
2
v
e
2
√
2
v
e
Validate
Solution:
If
v
be the velocity in the inter planetary space and
U
i
,
U
f
initial and final PE
Then, by the law of conservation of energy,
U
i
+
K
i
=
U
f
+
K
f
−
G
M
e
m
R
e
+
1
2
m
(
3
v
e
)
2
=
0
+
1
2
m
v
2
−
G
M
e
R
e
+
9
2
v
e
2
=
1
2
v
2
⇒
−
2
G
M
e
R
e
+
9
V
e
2
=
v
2
⇒
−
2
g
R
e
2
R
e
+
9
v
e
2
=
v
2
[
∵
G
M
e
=
g
R
e
2
]
−
2
g
R
e
+
9
×
2
=
g
R
e
=
v
2
,
[
∵
v
e
2
=
2
g
R
e
]
16
g
R
e
=
v
2
8.2
g
R
e
=
v
2
⇒
v
=
√
8
.
√
2
g
R
e
=
2
√
2
v
e
© examsnet.com
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