Calculating Escape Velocity on the Equator: Is It Possible?

[blost]

hello... I've got some small problem with Escape velocity. So... is it dependent on geographic coordinate ? I suppose "Yes!"... but is it true?

Using conservation of energy, I calculate this velocity on equator this way:

$$\frac{mv^2}{2}= \frac{GMm}{R} - \frac{m(2 \pi R)^2}{T^2*R }$$
Am I right ?

 

[Andrew Mason]

hello... I've got some small problem with Escape velocity. So... is it dependent on geographic coordinate ? I suppose "Yes!"... but is it true?

Using conservation of energy, I calculate this velocity on equator this way:

$$\frac{mv^2}{2}= \frac{GMm}{R} - \frac{m(2 \pi R)^2}{T^2*R }$$
Am I right ?

Are you trying to take into account the speed of the escaping object when it is sitting on the earth? Why?

To escape the earth, the object has to have enough kinetic energy so that its total energy (kinetic + potential) is greater than 0.

Condition for escape: KE + PE > 0.

AM

 

[RoyalCat]

hello... I've got some small problem with Escape velocity. So... is it dependent on geographic coordinate ? I suppose "Yes!"... but is it true?

Using conservation of energy, I calculate this velocity on equator this way:

$$\frac{mv^2}{2}= \frac{GMm}{R} - \frac{m(2 \pi R)^2}{T^2*R }$$
Am I right ?

You've assumed that your escaping mass is locked in a circular orbit around the earth, that is completely incorrect.
I also don't think you're comparing any actual energies of the object. What two states are you comparing? Try and compare the state where it's released from the surface of the earth, and the one when it has escaped it (It is no longer influenced by Earth's gravitational field, what does that mean energy-wise?)

Read Andrew Mason's pointers, they should help you get started, and you should then see whether the solution depends on your initial position or not.

 

[blost]

I know that KE+Pe>0.
And I know that escaping mass isn't locked in a circular orbit around the earth, but it's still got a kinetic energy from Earth rotation, so i don't know, is Kinetic energy a sum of kinetic energy of Earth rotation and energy which must be inserted to this object?

 

[Andrew Mason]

I know that KE+Pe>0.
And I know that escaping mass isn't locked in a circular orbit around the earth, but it's still got a kinetic energy from Earth rotation, so i don't know, is Kinetic energy a sum of kinetic energy of Earth rotation and energy which must be inserted to this object?

Yes, of course. But the question asks for the escape velocity, not the velocity or energy that you must add in order to achieve the escape velocity.

The reason the equator is important in this question is this: the distance from the Earth's surface to the centre varies with latitude. It is greatest at the equator.

AM

 

[blost]

Yes, of course. But the question asks for the escape velocity, not the velocity or energy that you must add in order to achieve the escape velocity.

thanks. Now i know...

by the way... sorry for my not-recherche language. English is not my native language (I'm 17 from Poland-3 h eng for a week ;/ 37 people in my class ;/ as like Cambodia... ), so please be tolerate :)