Escape velocity poles vs. equator.

[Soccerdoc11]

OK question from Kaplan and their explanation:

The fact that the Earth is rotating about its polar axis affects the escape velocity from the surface of the planet. Taking into account the Earth's rotation, the escape velocity at the North Pole is:

greater than the escape velocity at the South Pole.

less than the escape velocity at the South Pole.

greater than the escape velocity at the equator.

less than the escape velocity at the equator.

To answer this question consider how the Earth's rotation affects the force on a particle on the Earth's surface. Since the Earth is rotating, a particle on its surface exhibits uniform circular motion at a radius of r, which is equal to the distance of the particle from the axis of rotation or the polar axis. Centripetal acceleration is the translational velocity squared, divided by the radius of curvature for the translational path. The centripetal force is therefore F = mv2/r , where m is the mass of the particle and v is the speed of the particle. The speed of the particle is equal to the angular frequency with which the Earth rotates ω times by the distance between the particle and the axis of rotation r. Substituting v = rω, gives F = mω2r. Since m and ω are constants of the motion, the centripetal force increases as r increases. Since r is the distance between a particle on the surface of the Earth and the polar axis, r is a maximum at the equator, and therefore the centripetal force is a maximum at the equator.
Remember, the centripetal force is the force required to keep a particle traveling in a circular orbit of radius r. In this case, the force of gravity is the centripetal force. Since the force of gravity is approximately the same for all particles on the surface of the Earth and since the particles at the poles require a smaller attractive force to keep them at the surface of the Earth than the particles at the equator, the velocity required to escape from the surface of the Earth will be greater at the poles than at the equator, and choice C is correct.




I don't understand their explanation, but using simply Fc=mv^2/r wouldn't it essentially mean that the further you are then less force and easier to escape so less velocity necessary which is why at the poles it is tougher because the distance is smaller to the axis. I don't get their use of the angle frequency and the distance they use with the new equation. "Since the force of gravity is approximately the same for all particles on the surface of the Earth and since the particles at the poles require a smaller attractive force to keep them at the surface of the Earth than the particles at the equator, the velocity required to escape from the surface of the Earth will be greater at the poles than at the equator" This doesn't seem correct to me. Any help would be appreciated.. thanks

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[engineeredout]

Actually since the earth is an ellipse, it means that it is wider in the middle, therefore the distance from the center at the north pole would be less than the distance from the center at the equator, meaning the force of gravity would be greater.

Thats how I would think about it.

 

[Soccerdoc11]

Yeah the force of gravity would be greater so larger velocity is needed to overcome the force.. That's how I was thinking about it but just got confused with all that "theory" they used in the answer.. Thanks

 

[sleepy425]

OK question from Kaplan and their explanation:

The fact that the Earth is rotating about its polar axis affects the escape velocity from the surface of the planet. Taking into account the Earth's rotation, the escape velocity at the North Pole is:

greater than the escape velocity at the South Pole.

less than the escape velocity at the South Pole.

greater than the escape velocity at the equator.

less than the escape velocity at the equator.

To answer this question consider how the Earth's rotation affects the force on a particle on the Earth's surface. Since the Earth is rotating, a particle on its surface exhibits uniform circular motion at a radius of r, which is equal to the distance of the particle from the axis of rotation or the polar axis. Centripetal acceleration is the translational velocity squared, divided by the radius of curvature for the translational path. The centripetal force is therefore F = mv2/r , where m is the mass of the particle and v is the speed of the particle. The speed of the particle is equal to the angular frequency with which the Earth rotates ω times by the distance between the particle and the axis of rotation r. Substituting v = rω, gives F = mω2r. Since m and ω are constants of the motion, the centripetal force increases as r increases. Since r is the distance between a particle on the surface of the Earth and the polar axis, r is a maximum at the equator, and therefore the centripetal force is a maximum at the equator.
Remember, the centripetal force is the force required to keep a particle traveling in a circular orbit of radius r. In this case, the force of gravity is the centripetal force. Since the force of gravity is approximately the same for all particles on the surface of the Earth and since the particles at the poles require a smaller attractive force to keep them at the surface of the Earth than the particles at the equator, the velocity required to escape from the surface of the Earth will be greater at the poles than at the equator, and choice C is correct.




I don't understand their explanation, but using simply Fc=mv^2/r wouldn't it essentially mean that the further you are then less force and easier to escape so less velocity necessary which is why at the poles it is tougher because the distance is smaller to the axis. I don't get their use of the angle frequency and the distance they use with the new equation. "Since the force of gravity is approximately the same for all particles on the surface of the Earth and since the particles at the poles require a smaller attractive force to keep them at the surface of the Earth than the particles at the equator, the velocity required to escape from the surface of the Earth will be greater at the poles than at the equator" This doesn't seem correct to me. Any help would be appreciated.. thanks

well, the thing is, the question has nothing to do with the fact that earth is a spheroid (slightly elliptical). It's all about the centripital force at the pole versus at the equator.

At the equator:
the perpendicular distance from the equator to the axis of rotation is large. in other words, if you drew a perpendicular line from the axis to the earth's surface at the equator, it would be 6378.1 km.

At either pole:
the perpendicular distance from the equator to the axis of rotation is zero. in other words, if you drew a perpendicular line from the axis to the earth's surface at the pole, it would be 0 km.

now let's look at centripetal force:

at the equator:
centripetal force is large (Fc = mω2r and angular velocity is the same at all points on the earth because the earth is a solid body and so at all points the rotations per unit time are the same and we already established that r is large so Fc is large)

at the poles:
centripetal force is 0 (Fc= mω2r and angular velocity is the same at all points, but r is zero as we said above, so Fc is 0)

ok, now what does centripetal force mean? it's the force needed to keep the object on the surface of the body. it's NOT the force with which the earth pulls on the object, it's just the amount of force that's needed to keep the object on the surface.

ok, so now the earth pulls on all objects on the surface with pretty much the same force. sure, it's slightly smaller at the poles, but not significantly.

so if the earth pulls with the same force, but at the poles it takes less force to keep the object on the surface, then it will take higher velocities at the pole to get away from the surface. in other words, escape velocity is just the velocity that is needed to overcome the difference between the force of earth's pull and the centripetal force.

so basically at the poles, the earth pulls the same, but it's easier for the earth to hold the object there, so it's harder (takes higher velocity) for the object to escape. at the equator, it's harder for the earth to hold the object, but it still pulls the same, so it's easier (takes lower velocity) for the object to fly off.

 

[Soccerdoc11]

so basically at the poles, the earth pulls the same, but it's easier for the earth to hold the object there, so it's harder (takes higher velocity) for the object to escape. at the equator, it's harder for the earth to hold the object, but it still pulls the same, so it's easier (takes lower velocity) for the object to fly off.

Perfetto.... Thanks sleepy - the wording was killing me but that's much clearer

 

[sleepy425]

no prob