Recent content by AakashPandita

Wow, Thanks a lot! I think I understand it fully now. Thanks again!

So the original thing that we launched a body upwards with a speed ##2\sqrt{gR} ## so that it starts to orbit the Earth at ## \sqrt{2gR} ## is wrong ?

## 1/2 m v^2 - (GmM) / r =0 ## This is the condition for escape velocity . Right? So if it has escape velocity and it is continuously moving away from the earth, r will keep increasing. So its velocity should decrease with r and it should come to a stop at some point?

Then how do we make things orbit?

why wouldn't it orbit?

If it has escape velocity, why would it escape ? Writing error?

Then how do we increase the speed of any orbiter?

If the launching speed is more than escape velocity, doesn't the extra energy add to the KE of body in orbit, i.e velocity? A body can orbit if its kinetic energy is equal to its gravitational potential energy. What happens if it has even more energy?

The given initial speed was ## 2\sqrt{gR}##.In the second attachment I obtained the orbit speed.

Why can't we consider it to be launched in circular orbit like a satellite? The orbit assumed to be circular?

I have taken h to be the distance from the center. Then how can it be less than R?

But the object isn't revolving at the given initial velocity.

Homework Statement There was a question in which an object was released upwards from Earth with a velocity of ## 2\sqrt {gR}##. Using conservation of energy I found that the speed of the body in the orbit was ## \sqrt {2gR} ## Everything's fine till now. But then I wanted to find the height at...

I got the answer. Thanks a lot.

Wait a minute. Doesn't the value of g that we use account for the centripetal force due to rotation of earth? Is ## g- (\omega)^2 R =g ## using approximation? If yes then how?