Answer and Solution:
As per the question:
Mass of rocket = m
Distance from the center = R
Now,
(a) For tangential impulse, the centripetal force of the rocket is provided by the gravitational force, thus:
[tex]F_{C} = F_{G}[/tex]
where
[tex]F_{C}[/tex] = Centripetal force
[tex]F_{G}[/tex] = Gravitational force
Therefore,
[tex]\frac{mv^{2}}{R} = \frac{GMm}{R^{2}}[/tex]
[tex]v = \sqrt{\frac{GM}{R}}[/tex]
Also, the velocity of the rocket after impulse is given as:
[tex]mv = m(v + \del v)[/tex]
[tex]v = v + \del v[/tex]
Now, the tangential impulse can be calculated as under:
At infinity, the energy is zero:
[tex]\frac{1}{2}m(v + \del v)^{2} = \frac{GM}{R}[/tex]
[tex]v + \del v = \sqrt{\frac{2GM}{R}}[/tex]
[tex]\del v = \sqrt{\frac{2GM}{R}} - v[/tex]
[tex]\del v = \sqrt{\frac{2GM}{R}} - \sqrt{\frac{GM}{R}} = \sqrt{\frac{GM}{R}}[/tex]
Now,
Impulse, [tex]m\del v = m\sqrt{\frac{GM}{R}}[/tex]
(b) Resulting orbit will follow a straight line path.
(c) On comparing the radial impulse, the energy that the impulse has provided is given by:
[tex]\frac{1}{2}mv^{2} = \frac{GmM}{R}[/tex]
[tex]v = \sqrt{\frac{2GM}{R}}[/tex]
Impulse = mv = [tex]m\sqrt{\frac{2GM}{R}}[/tex]
