Physics ? from Kaplan FL5

[sjhalverson]

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I don't understand their reasoning on this one:


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:
A greater than the escape velocity at the South Pole.
B less than the escape velocity at the South Pole.
C greater than the escape velocity at the equator.
D 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 ? times by the distance between the particle and the axis of rotation r. Substituting v = r?, gives F = m?2r. Since m and ? 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 chose D for the same reason they chose C. If particles at the equator have a greater centripetal force pulling them in to the earth (when looking at the earth from the north pole), shouldn't they require a greater velocity to overcome that force?

I feel like "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" is a contradiction

 

[slyloxy]

I don't understand their reasoning on this one:


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:
A greater than the escape velocity at the South Pole.
B less than the escape velocity at the South Pole.
C greater than the escape velocity at the equator.
D 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 ? times by the distance between the particle and the axis of rotation r. Substituting v = r?, gives F = m?2r. Since m and ? 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 chose D for the same reason they chose C. If particles at the equator have a greater centripetal force pulling them in to the earth (when looking at the earth from the north pole), shouldn't they require a greater velocity to overcome that force?

I feel like "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" is a contradiction

Escape velocity v=sqrt (GM/r) v is inverse proportional to r and is independent of the mass of the particle. if r is bigger at the equator then is escape velocity will be smaller. if r is smaller in the case at the poles then escape velocity will larger. is easier to look it this way. if the centripetal force is huge then it is hard to attain very huge velocities. most of the energy input has to be used to overcome the earths gravititational potential and very small can be used to provide kinetic energy and hence lower velocity to escape the earth

 

[slyloxy]

Escape velocity v=sqrt (GM/r) v is inverse proportional to r and is independent of the mass of the particle. if r is bigger at the equator then is escape velocity will be smaller. if r is smaller in the case at the poles then escape velocity will larger. is easier to look it this way. if the centripetal force is huge then it is hard to attain very huge velocities. most of the energy input has to be used to overcome the earths gravititational potential and very small can be used to provide kinetic energy and hence lower velocity to escape the earth

escape velocity v=sqrt(2GM/r) my bad forgot the sqrt 2