Are you trying to take into account the speed of the escaping object when it is sitting on the earth? Why?blost said: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:
[tex]\frac{mv^2}{2}= \frac{GMm}{R} - \frac{m(2 \pi R)^2}{T^2*R }[/tex]
Am I right ?
blost said: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:
[tex]\frac{mv^2}{2}= \frac{GMm}{R} - \frac{m(2 \pi R)^2}{T^2*R }[/tex]
Am I right ?
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.blost said: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 said: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.
Escape velocity is the minimum speed needed for an object to escape the gravitational pull of a larger object, such as a planet or moon. It is calculated using the formula v = √(2GM/r), where G is the universal gravitational constant, M is the mass of the larger object, and r is the distance from the center of the object to the center of the smaller object.
Yes, it is possible to calculate escape velocity on the equator. The equator is simply a location on the surface of a planet or moon, so the same formula for escape velocity can be used.
The rotation of the Earth does not have a significant effect on the escape velocity on the equator. While the Earth's rotation does affect the overall velocity of an object, the effect is very small compared to the large mass of the Earth.
No, escape velocity on the equator is the same as other locations on Earth. The difference in distance from the center of the Earth to the center of the object is negligible, so the same escape velocity formula can be used.
Yes, other factors such as the atmospheric conditions and the presence of other objects near the equator can affect the escape velocity. However, these effects are typically small and can be accounted for in the calculations.