Question

A meteorite approaching a planet of mass M(in the straight line passing through the centre of the planet) collides with automatic space station orbiting the planet in a circular trajectory of radius R.The mass of the station is ten times as large as the mass of meteorite.As a result of the collision,meteorite sticks with the station which goes over to a new orbit with minimum distance R/2 from the planet.Speed of meteorite just before it collides with the space station is

Answer 1

Hi <3

Answer 2

I made a hit upon this bt what is confusing me is the statement minimum distance as R/2.

Answer 3

Don't know how to use it in this problem!

Answer 4

One thing's for sure: we can't do energy conservation here, 'cuz it's an inelastic collision.

Answer 5

We can apply angular momentum conservation at the centre of the orbit.
(That is planet)

Answer 6

Let's look at it in terms of angular momentum?\[10mvR = 11mv'(R/2)\]

Answer 7

R/2 is the minimum distance .
How did u generalized it for any part of orbit?

Answer 8

Exactly, haha.
I guess we also know \(v\) which is the orbital velocity.
And I'll explain that in the next post. Hold on a sec.

Answer 9

v is √GM/R

Answer 10

Do you agree that whatever kind of orbit we're looking at, we could say this: at the minimum distance, the velocity is perpendicular to the position vector. Try to imagine what happens otherwise.

Answer 11

And of course angular momentum is conserved at each point of the orbit, as gravity does not apply any torque about the center of the planet!

Answer 12

Yes angular momentum is conserved for each point of orbit but are u just taking it for the minimum distance becoz we know only that .
Lol

Answer 13

The real purpose of telling us the "minimum distance" part is that \(\theta = \pi/2\) and that \(mvr \sin \theta = mvr\).

Answer 14

Okay!

Answer 15

Well, actually we can conserve energy. But for that, we can only look at situations AFTER the collision.

Answer 16

So can you guess the second equation now? :D

Answer 17

I will sum up the solution:

1. Use angular momentum conservation. Now we know the velocity at the point where the distance is R/2.

2. Use linear momentum conservation for just before and just after the collision. Assume a variable for velocity of the meteorite (which we ultimately have to find) say \(x\). Keep in mind that velocity of the station and meteorite are perpendicular to each other!
From 2, you will have the velocity immediately after the collision in terms of \(x\). The distance from the planet is \(R\).

3. Now equate energies from immediately after the collision and the point where the distance is R/2.

Good luck =)

Answer 18

I will try it !
If i face a difficulty would you please help ?

Answer 19

that's what I'm here for, lol

Answer 20

One doubt in my mind.
You changed your first statement here!
(Statement was we won't be able to apply energy conservation).
You falsified it in the next statement of yours.So ironical!Would you justify it?

Answer 21

What I meant from my initial statement was this: we can't say that energy before the collision is the same as energy after the collision.
However, once the collision has taken place, we can choose any two times to equate energy.

Now we're gonna use energy conservation at these two times:
One, immediately after the collision.
Two, at the time when the distance is R/2.

Clearly, both are after the event of the collision.

Answer 22

Okay..
Parth listen.
It got this v'=20v/11
And v'=√x^2+100v^2/121
So we can simply equate the two to get the ans.
We know v also ..
So this can work out.

Answer 23

v is √GM/R

Answer 24

No use of energy conservation. Isn't it?

Answer 25

the v' = 20v/11 is at the time when the i