Another centripetal forces problem

[thechairman]

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66. 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.

The answer is C. But I really don't see why. I know that gravity mediates the centripetal force of the hypothetical object located the surface. As latitude increases, the distance to the center of axis decreases, so the centripetal force required to keep the object on spinning on earth, decreases. But gravitational pull stays the same. I don't know how to use this information to figure out escape velocity. Shouldn't the escape velocity be lower at the poles than the equator because the centripetal force required decreases?


Thanks

Sorry to ask so many questions here. man i feel like i'm freeloading off you guys. but i promise i'll contribute!

 

[chad5871]

This could be totally incorrect and wrong, but this is how I would reason this one out. If it's wrong, expect me to come back and delete this post when someone posts the correct logic. 😛

This question is dealing with Newton's first law, "an object in motion tends to remain in that state of motion unless acted upon by an external force." The object at the north pole will have no velocity because there is no displacement. The object stays in the same place in space. The object at the equator, however, has centripetal acceleration because, although the magnitude of the velocity is constant, the direction is constantly changing. Also remember that velocity in this case is at a 90 degree angle to the rotational axis. Thus, the object at the equator will want to stay in that state of motion. Because the escape velocities are the same when the Earth is not rotating, this initial velocity on the object at the equator will cause the escape velocity to be lessened.

 

[wcbcruzer]

It's greater at the north pole because there is centrifugal force pushing an object outward (away from Earth) at the equator, but none at the north pole. So it takes more "escape velocity" for something to be propelled away from Earth at the North pole.

 

[chad5871]

It's greater at the north pole because there is centrifugal force pushing an object outward (away from Earth) at the equator, but none at the north pole. So it takes more "escape velocity" for something to be propelled away from Earth at the North pole.

Yeah, that's pretty much what I meant to say, but wcbcruzer just said it much more succinctly and clearly. 👍

 

[Nowaythisnameis]

66. 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.

The answer is C. But I really don't see why. I know that gravity mediates the centripetal force of the hypothetical object located the surface. As latitude increases, the distance to the center of axis decreases, so the centripetal force required to keep the object on spinning on earth, decreases. But gravitational pull stays the same. I don't know how to use this information to figure out escape velocity. Shouldn't the escape velocity be lower at the poles than the equator because the centripetal force required decreases?


Thanks

Sorry to ask so many questions here. man i feel like i'm freeloading off you guys. but i promise i'll contribute!

could u post the explanation provided by the test company? the "centrifugal force" argument is weak at best because there is no such thing as centrifugal force. It seems like the answer should be D. Are u sure the answer is C? And if it is C, it would really help if you posted what the testing company said about it.

 

[pyromatic]

The angular velocity of the earth is fixed since it makes one full rotation approximately every 24 hours. So your linear velocity depends only on your distance from the center of rotation. Imagine you squished the earth so that the north and south poles came together; the earth is now a disc with the equator at the very edge and the poles at the center. Someone standing at the equator will have a greater linear velocity than someone standing halfway out who will have a greater linear velocity than someone standing at the center (north pole) who will have a linear velocity of zero. According to google, the diameter of the earth is 12756.1 km, so someone standing at the equator is already traveling at about 460 m/s, and since the escape velocity is a constant 11.2 km/s on the surface of the earth, they have a head start.

 

[wcbcruzer]

could u post the explanation provided by the test company? the "centrifugal force" argument is weak at best because there is no such thing as centrifugal force. It seems like the answer should be D. Are u sure the answer is C? And if it is C, it would really help if you posted what the testing company said about it.

I know there's no such thing as centrifugal force, but it's a good way to explain the problem.

 

[changarang]

i don't know if this will help with the visualization, but if you imagine being on a merry-go-round that's spinning really fast, does it make sense that you'd fly off more easily the farther out you are from the center? it's for the reasons already stated...objects at the equator are already moving with a linear velocity equal to the circumference of the equator per amount of time, whereas objects at the poles have a linear velocity of zero (because they're a distance of zero from the axis or center of rotation). since objects at the poles are starting from a velocity of zero, it's harder for them to reach the escape velocity.

 

[chad5871]

i don't know if this will help with the visualization, but if you imagine being on a merry-go-round that's spinning really fast, does it make sense that you'd fly off more easily the farther out you are from the center? it's for the reasons already stated...objects at the equator are already moving with a linear velocity equal to the circumference of the equator per amount of time, whereas objects at the poles have a linear velocity of zero (because they're a distance of zero from the axis or center of rotation). since objects at the poles are starting from a velocity of zero, it's harder for them to reach the escape velocity.

Oooh, good illustration! I was trying to think of something clever to explain this concept, and you hit the nail on the head! 👍

 

[Genie133]

i don't know if this will help with the visualization, but if you imagine being on a merry-go-round that's spinning really fast, does it make sense that you'd fly off more easily the farther out you are from the center? it's for the reasons already stated...objects at the equator are already moving with a linear velocity equal to the circumference of the equator per amount of time, whereas objects at the poles have a linear velocity of zero (because they're a distance of zero from the axis or center of rotation). since objects at the poles are starting from a velocity of zero, it's harder for them to reach the escape velocity.

that's a really good explanation.... i didn't get that problem either (it looks like we just did the same test, chairman)