Question 45: Gravitation

Question

A rocket is launched normal to the surface of the Earth, away from the sun, along the line joining the Sun and the Earth. The Sun is $$3 	imes 10{}^5$$ times heavier than the earth and is at a distance  $$2.5 	imes {10^4}$$ times larger than the radius of the Earth. The escape velocity from Earth's gravitational field is $${V_c} = 11.2km\,{s^{ - 1}}.$$. The minimum initial velocity $$\left( {{v_s}} ight)$$  required for the rocket to be able to leave the sun-earth system is closest to (Ignore the the rotation and revoluation of the earth and the presence of any other planet)

Accepted Answer

Given : Mass of the sun, M
s
= 3 $$	imes$$ 10
5
$$	imes$$ mass of the earth = 3 $$	imes$$ 10
5
M
e
Distance between the sun and the earth,
d = 2.5 $$	imes$$ 10
4
$$	imes$$ radius of the earth = 2.5 $$	imes$$ 10
4
R
e
,
Escape speed, v
e
= 11.2 km s
$$-$$1
$$	herefore$$ $${v_e} = \sqrt {{{2G{M_e}} \over {{R_e}}}}  = 11.2$$ km s
$$-$$1
Minimum velocity required for the rocket to escape the given earth $$-$$ sun system = v
s
Using energy conservation for the given situation,
$${1 \over 2}mv_s^2 = {{G{M_e}m} \over {{R_e}}} + {{G{M_s}m} \over {d + {R_e}}} = {{G{M_e}m} \over {{R_e}}} + {{G 	imes 3 	imes {{10}^5}{M_e}m} \over {(2.5 	imes {{10}^4}{R_e} + {R_e})}}$$
$$	herefore$$ $$v_s^2 = {{2G{M_e}} \over {{R_e}}} + {{24G{M_e}} \over {{R_e}}}$$ ($$\because$$ 2.5 $$	imes$$ 10
4
>> 1)
$$ \Rightarrow v_s^2 = v_e^2 + 12v_e^2 = 13v_e^2$$
or $${v_s} = \sqrt {13} $$ v
e
= 40.38 km s
$$-$$1
$$\approx$$ 42 km s
$$-$$1

Answer

$${v_s} = 22\,km\,{s^{ - 1}}$$

Answer

$${v_s} = 42\,km\,{s^{ - 1}}$$

Answer

$${v_s} = 62km\,{s^{ - 1}}$$

Answer

$${v_s} = 72km{s^{ - 1}}$$

Gravitation

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