Finding height of revolving body at speed ##\sqrt{2gR}##

[AakashPandita]

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 which the body was traveling from the centre of the earth.

Homework Equations

Equating centripetal force and gravitational force
v is the speed of body
m is mass of body.
R is radius of earth.
The body is at a height of h from the center of the earth.
To find h.

The Attempt at a Solution

h is coming out to be less that R. How is that possible?

 

[PeroK]

I don't know how you worked out the orbital height or velocity.

Try to work out the orbital height for the given initial velocity. I think you'll see that there's something wrong with the question.

 

[cnh1995]

I haven't done any calculations but h can be less than R..R is the radius of the Earth (distance from center to surface) and h is the height at which the object is above the surface...

 

[AakashPandita]

I don't know how you worked out the orbital height or velocity.

Try to work out the orbital height for the given initial velocity. I think you'll see that there's something wrong with the question.

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

 

[AakashPandita]

I haven't done any calculations but h can be less than R..R is the radius of the Earth (distance from center to surface) and h is the height at which the object is above the surface...

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

 

[BvU]

Dear Ap,

Where does your first equation come from ?
To me it looks as if it comes from assuming a circular orbit around the earh.
But from the problem statement I get the impression the object is sent off in a vertical direction. With an enormous speed. It can get very far away from Earth ! How far ? Your turn !

 

[AakashPandita]

Dear Ap,

Where does your first equation come from ?
To me it looks as if it comes from assuming a circular orbit around the earh.
But from the problem statement I get the impression the object is sent off in a vertical direction. With an enormous speed. It can get very far away from Earth ! How far ? Your turn !

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

 

[BvU]

Actually, that was part II of my approach.

But let's not go too fast and follow your assumption: the object is launched with a given speed.

In order to bring the object from where it starts off to the height of its 'orbit', it must loose some of its kinetic energy to overcome the gravitational attraction by the earth. In your first equation you use the same speed as the object had at launch, so this trade-in of kinetic energy for potential energy from gravity isn't taken into account. Can you think of a relationship that does handle that a little better ?

 

[AakashPandita]

Actually, that was part II of my approach.

But let's not go too fast and follow your assumption: the object is launched with a given speed.

In order to bring the object from where it starts off to the height of its 'orbit', it must loose some of its kinetic energy to overcome the gravitational attraction by the earth. In your first equation you use the same speed as the object had at launch, so this trade-in of kinetic energy for potential energy from gravity isn't taken into account. Can you think of a relationship that does handle that a little better ?

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

 

[BvU]

Oh, sorry. I read over that $2\sqrt{gR}$ altogether.

But don't you get a little suspicious if the launching speed is $\sqrt 2$ times the escape velocity ? What induces you to think there will be an orbit at all ?

 

[AakashPandita]

Oh, sorry. I read over that $2\sqrt{gR}$ altogether.

But don't you get a little suspicious if the launching speed is $\sqrt 2$ times the escape velocity ? What induces you to think there will be an orbit at all ?

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?

 

[BvU]

As you calculated, it loses half its kinetic energy to escape from the influence of the Earth gravitational field. If there are no other celestial bodies (sun !?) around (*) to play a role, the remaining kinetic energy is constant and the thing flies off into infinity at the escape velocity, 11.2 m/s

The "part II" I alluded to is that it doesn't matter in which direction the launch takes place (as long as it's upward ), because it's a calculation with energies.

(*) there is also the rotation of the Earth around its axis to take into account, as well as the rotation of the Earth around the sun, and there is some friction with the atmosphere, but we seem to be allowed to ignore those here.

 

[AakashPandita]

As you calculated, it loses half its kinetic energy to escape from the influence of the Earth gravitational field. If there are no other celestial bodies (sun !?) around (*) to play a role, the remaining kinetic energy is constant and the thing flies off into infinity at the escape velocity, 11.2 m/s

The "part II" I alluded to is that it doesn't matter in which direction the launch takes place (as long as it's upward ), because it's a calculation with energies.

(*) there is also the rotation of the Earth around its axis to take into account, as well as the rotation of the Earth around the sun, and there is some friction with the atmosphere, but we seem to be allowed to ignore those here.

Then how do we increase the speed of any orbiter?

 

[BvU]

First of all, correct my 11.2 m/s to 11.2 km/s. A mere factor 1000 .

Increasing the speed of an orbiter ? You don't mean something like "by accelerating the thing, e.g. with a rocket engine" , I suppose.

There's the orbit itself to consider: for circular orbits there are simple expressions. Once in a circular orbit, accelerating)/decelerating) causes the orbit to become elliptical (Kepler orbit), i.e. initially the thing goes higher or lower. Proper acceleration/deceleration at a later point can make the orbit circular again.

 

[AakashPandita]

If there are no other celestial bodies (sun !?) around (*) to play a role, the remaining kinetic energy is constant and the thing flies off into infinity at the escape velocity, 11.2 m/s

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

 

[BvU]

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

I don't understand this question. If it has escape velocity, it has enough kinetic energy to overcome the graviational field. So it can go infinitely far.

 

[AakashPandita]

why wouldn't it orbit?

 

[BvU]

The attracting force from the Earth isn't enough to curve the trajectory sufficiently.
Or: by the time the radial deceleration has dropped to zero, there is still outward radial speed left over.

 

[AakashPandita]

Then how do we make things orbit?

 

[BvU]

Launching them with a lower speed !
(Actually, speed isn't the issue: energy is what counts -- as we have seen )
11.2 km/s is awfully fast (33 x speed of sound) !

 

[AakashPandita]

I don't understand this question. If it has escape velocity, it has enough kinetic energy to overcome the graviational field. So it can go infinitely far.

$1/2 m v^2 - (GmM) / r =0$
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?

 

[BvU]

In a way, yes. From your expression: v = 0 by the time r is so big that GMm/r is also 0. So pretty much: at infinity !

 

[AakashPandita]

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 ?

 

[BvU]

I haven't seen the actual question formulation, but the way you put it: yes, there is considerable overshoot so calling it an orbit in the usual sense of a curved trajectory around the Earth isn't right.

 

[AakashPandita]

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