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A projectile is ejected with vertical speed $v$ from the surface of a planet of mass $M$ and radius $R$

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A projectile is ejected with vertical speed $v$ from the surface of a planet of mass $M$ and radius $R$. Show that it comes to rest at a distance r from the centre of the planet where $$r=R/(1-Rv^2/2GM)$$ If $ v = c$ in the above example what will be the condition between M and R such that the planet acts like a Newtonian black hole.

I have tried to solve it in the following way-

The height from the surface of the planet will be given by-
$$y=vt-(1/2)gt^2$$
when it comes to rest $v=0$ hence from the above equation we get
$$r-R=(1/2)gt^2$$
From relation between g and G,
$$g=Gm/r^2$$
I have put all this values in $$y=vt-(1/2)gt^2$$
but I am not getting the required answer.

Please tell me where did I made the mistake and correct me if going in the wrong direction. Thanking you.

asked Dec 9, 2018 in Physics Problems by GB (120 points)
edited Dec 15, 2018 by sammy gerbil
Please show your attempt. Or explain your difficulty.

1 Answer

1 vote

Your mistake is that you assumed that the acceleration due to gravity $g$ is constant. This is only approximately true over distances which are very much smaller than the average distance of the object from the centre of the planet. The equation $y=vt-\frac12 gt^2$ only applies for constant acceleration $g$.

You need to use the conservation of energy. The initial KE plus gravitational PE at the surface of the planet is equal to the final KE plus gravitational PE at any other point such at that where it comes to rest instantaneously.

If the projectile were not fired vertically but instead at an angle to the vertical it would be necessary also to apply the conservation of angular momentum. In this problem the angular momentum is always zero.

If $v=c$ then Special Relativity Theory ought to apply instead of Newtonian Mechanics, so the kinetic energy would not be $\frac12 mv^2$. However this question assumes that Newtonian Mechanics is still valid when $v=c$. The tags for Special and General Relativity are not needed for this question.

answered Dec 16, 2018 by sammy gerbil (28,746 points)
edited Dec 16, 2018 by sammy gerbil
Thank you so much.
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