Question

You are in outer space at point A and you are “falling” towards Jupiter. But lucky for you, you’re “falling” at a speed of ‘v’ which is higher than the escape velocity of Jupiter’s gravitational field at a distance ‘R’ as shown in the figure. Your prayers to the Old Gods and the New seem to be working because luck is on your side once again. You don’t fall directly onto the planet but go around it and away from it because of your escape velocity. When you reach point B, what is your speed(v’) ? (Assume the only force that acted on you is Jupiter’s gravitational pull. Also assume Jupiter is stationary.)

• A. $v$
• B. $2v$
• C. $\sqrt{v}$
• D. None of these

Answer 1

The correct option is A $v$

Ok. So the only force acting on you is Jupiter’s gravity. That means you are in Jupiter’s gravitational field. Now Gravitational field is a conservative field (just like electrostatic field), which means the energy of a mass remains conserved in such a field.
Now you have a kinetic energy because of your velocity ‘v’ at A and a potential energy due to Jupiter’s gravity at A. The same goes for the point B.
Now we know,
$(KE+PE{)}_A=(KE+PE{)}_B$
$P{E}_A=GMm/R$
$P{E}_B=GMm/R$
We see that $P{E}_A=P{E}_B$
That must mean $K{E}_A=K{E}_B$
Thus, $\frac{m{v}^2}{2}=\frac{m{v}^{{}^{\prime }2}}{2}$
$v^{{}^{\prime }}=v$

Now maybe if you had been praying to the Lord of Light you wouldn’t have been stuck floating around in space.