a rocket of mass 0.200 kg is launched from rest of reaches point p lying at a height of 30m above the surface of the earth from the starting point in the process +425j of work is done on the rocket by burning chemical propellant ignoring air resistance and the amount of mas lost due to the burning propellant find the speed of the rocket at point p
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Answer:
Explanation:
We can get the answer by energy conservation laws
Total energy at surface = total energy at the highest point
- GMm/ R +1/2mv2= GMm/(R+h) +0
Now get h
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Mass of the Sun, M
s
=2×10
30
kg
Mass of the Earth, M
e
=6×10
24
kg
Orbital radius, r=1.5×10
11
m
Mass of the rocket =m
Let x be the distance from the centre of the Earth where the gravitational force acting on satellite P becomes zero.
From Newton’s law of gravitation, we can equate gravitational forces acting on satellite P under the influence of the Sun and the Earth as:
(r−x)
2
GmM
s
=
x
2
GmM
e
[
x
(r−x)
]
2
=M
s
/M
e
(r−x)/x=(
60×10
24
2×10
30
)
1/2
=577.35
1.5×10
11
−x=577.35x
578.35x=1.5×10
11
x=1.5×10
11
/578.35=2.59×10
8
m.
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