## ParthKohli one year ago Another nice physics question.

1. ParthKohli

A space vehicle is approaching a planet with a speed $$v$$ and is a long way out (treat it as infinity). The trajectory would miss the centre of the planet by distance $$R$$ if it continued in a straight line. If the planet has a mass $$m$$ and a distance $$r$$, what is the smallest value of $$R$$ in order that the resulting orbit will just miss the surface?

2. ParthKohli

So I'll try using conservation of energy here probably.|dw:1441717901150:dw|

3. ParthKohli

|dw:1441718086223:dw|

4. Abhisar

Is r the radius of the planet?

5. ParthKohli

$\rm{Initial ~energy:}~~\frac{1}{2}mv^2$Now the gravitational force does some work to bring it from its original trajectory to the surface. This work is equal to the difference in potential energy.$= \frac{GMm}{r} - \frac{GMm}{R}$

6. ParthKohli

@Abhisar Yes.

7. ParthKohli

OK, sorry, my OS got completely hung.

8. ParthKohli

Alright. Ugh. I'll get to the point in that case.$\text{initial energy + change in potential energy = final energy}$Now,$\frac{1}{2}mv^2 + \frac{GMm}{r} - \frac{GMm}{R} = \frac{1}{2}m \left(\sqrt{GM/r}\right)^2 - \frac{GMm}{r}$

9. ParthKohli

$\frac{1}{2}mv^2 + \frac{GMm}{r} - \frac{GMm}{R} = -\frac{GMm}{2r}$$\Rightarrow \frac{1}{2}v^2 + \frac{3}{2r}GM = \frac{GM}{R}$

10. IrishBoy123
11. IrishBoy123

i'd build some DE's to solve this.