A rocket is fired vertically with a speed of 5
km 5-1 from the earth's surface. How far
from the earth does the rocket go before
returning to the earth? Mass of the earth =
6.0 x 1024 kg: mean radius of the earth =
6.4*106 m; G = 6.67 10-11 N m2 kg-2
(a) 7.33 x106m
(b) 7.33 x10m
(c) 9.356 x 10' m
(d) 9.356 x 10 m
Answers
Answer:
A rocket is fired vertically with a speed of 5 kms
−1
from the Earth's surface. How far from the Earth does the rocket go before returning to the Earth ? Mass of the earth =6.0×10
24
kg; mean radius of the radius of the Earth =6.4×10
6
m; G=6.67×10
−11
Nm
2
kg
−2
.
Answer:
A rocket is fired vertically with a speed of 5 kms
−1
from the Earth's surface. How far from the Earth does the rocket go before returning to the Earth ? Mass of the earth =6.0×10
24
kg; mean radius of the radius of the Earth =6.4×10
6
m; G=6.67×10
−11
Nm
2
kg
−2
.
Explanation:
Velocity of the rocket, v=5×10
3
m/s
Mass of the Earth, M
e
=6×10
24
kg
Radius of the Earth, R
e
=6.4×10
6
Height reached by rocket mass m=h
At the surface of the Earth,
Total energy of the rocket = Kinetic energy + Potential energy
=
2
1
mv
2
+(−
R
e
GM
e
m
)
At the highest point h,
v=0⟹Kinetic energy=0
Potential energy=−
R
e
+h
GM
e
m
Total energy of the rocket =−
R
e
+h
GM
e
m
By law of conservation of energy, total energy at surface = total energy at height h
2
1
mv
2
−
R
e
GM
e
m
=−
R
e
+h
GM
e
m
v
2
/2=
R
e
(R
e
+h)
GM
e
h
Substituting g=GM
e
/R
e
2
and rearranging the terms,
h=
2gR
e
−v
2
R
e
v
2
=
2×9.8×6.4×10
6
−(5×10
3
)
2
6.4×10
6
×(5×10
3
)
2
=1.6×10
6
m
Height achieved by the rocket with respect to the earth is
R
e
+h=6.4×10
6
+1.6×10
6
=8.010
6
m