- Thank you received: 0
Gravitons and Push Gravity question.
- tvanflandern
- Topic Author
- Offline
- Platinum Member
Less
More
19 years 11 months ago #12385
by tvanflandern
Replied by tvanflandern on topic Reply from Tom Van Flandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Jim</i>
<br />Anything falling into a hole through the Earth would accelerate for a while and then would be slowed down-right? So, where would it come to rest?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Barring some retarding force such as atmospheric drag, it would never come to rest. It would oscillate from one side of the Earth to the other and back again forever.
While we're having so much fun, consider this paradox. Special relativity says that fast motion causes time to slow. Earth's equator spins with a speed of about 0.5 km/s, or about the speed of a bullet. According to SR, time should pass more slowly on the speeding equator than at the stationary pole. The cumulative loss of time over the lifetime of the Earth is roughly 10,000 years. So if we start walking south from the pole in the year 2005, will we be visiting 10,000 years into Earth's past when we reach the equator? Why or why not? -|Tom|-
<br />Anything falling into a hole through the Earth would accelerate for a while and then would be slowed down-right? So, where would it come to rest?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Barring some retarding force such as atmospheric drag, it would never come to rest. It would oscillate from one side of the Earth to the other and back again forever.
While we're having so much fun, consider this paradox. Special relativity says that fast motion causes time to slow. Earth's equator spins with a speed of about 0.5 km/s, or about the speed of a bullet. According to SR, time should pass more slowly on the speeding equator than at the stationary pole. The cumulative loss of time over the lifetime of the Earth is roughly 10,000 years. So if we start walking south from the pole in the year 2005, will we be visiting 10,000 years into Earth's past when we reach the equator? Why or why not? -|Tom|-
Please Log in or Create an account to join the conversation.
- rousejohnny
- Offline
- Elite Member
Less
More
- Thank you received: 0
19 years 11 months ago #12028
by rousejohnny
Replied by rousejohnny on topic Reply from Johnny Rouse
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by tvanflandern</i>
<br /><blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Jim</i>
<br />Anything falling into a hole through the Earth would accelerate for a while and then would be slowed down-right? So, where would it come to rest?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Barring some retarding force such as atmospheric drag, it would never come to rest. It would oscillate from one side of the Earth to the other and back again forever.
While we're having so much fun, consider this paradox. Special relativity says that fast motion causes time to slow. Earth's equator spins with a speed of about 0.5 km/s, or about the speed of a bullet. According to SR, time should pass more slowly on the speeding equator than at the stationary pole. The cumulative loss of time over the lifetime of the Earth is roughly 10,000 years. So if we start walking south from the pole in the year 2005, will we be visiting 10,000 years into Earth's past when we reach the equator? Why or why not? -|Tom|-
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
No, because drag from the centrifugal force is "slowing" them down. Is that even a good guess?
<br /><blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Jim</i>
<br />Anything falling into a hole through the Earth would accelerate for a while and then would be slowed down-right? So, where would it come to rest?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Barring some retarding force such as atmospheric drag, it would never come to rest. It would oscillate from one side of the Earth to the other and back again forever.
While we're having so much fun, consider this paradox. Special relativity says that fast motion causes time to slow. Earth's equator spins with a speed of about 0.5 km/s, or about the speed of a bullet. According to SR, time should pass more slowly on the speeding equator than at the stationary pole. The cumulative loss of time over the lifetime of the Earth is roughly 10,000 years. So if we start walking south from the pole in the year 2005, will we be visiting 10,000 years into Earth's past when we reach the equator? Why or why not? -|Tom|-
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
No, because drag from the centrifugal force is "slowing" them down. Is that even a good guess?
Please Log in or Create an account to join the conversation.
- tvanflandern
- Topic Author
- Offline
- Platinum Member
Less
More
- Thank you received: 0
19 years 11 months ago #12033
by tvanflandern
Replied by tvanflandern on topic Reply from Tom Van Flandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by rousejohnny</i>
<br />No, because drag from the centrifugal force is "slowing" them down. Is that even a good guess?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">At least you took a guess. Good effort. But centrifugal force is not a form of drag.
The answer is that gravitational potential is weaker on the equator than at the poles, which makes clocks at the equator run faster. The two effects, clock-slowing from speed and clock speed-up from potential, exactly cancel, leaving Earth's entire surface as an "equichron" (equal time) surface. -|Tom|-
<br />No, because drag from the centrifugal force is "slowing" them down. Is that even a good guess?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">At least you took a guess. Good effort. But centrifugal force is not a form of drag.
The answer is that gravitational potential is weaker on the equator than at the poles, which makes clocks at the equator run faster. The two effects, clock-slowing from speed and clock speed-up from potential, exactly cancel, leaving Earth's entire surface as an "equichron" (equal time) surface. -|Tom|-
Please Log in or Create an account to join the conversation.
- cosmicsurfer
- Offline
- Platinum Member
Less
More
- Thank you received: 0
19 years 11 months ago #11019
by cosmicsurfer
Replied by cosmicsurfer on topic Reply from John Rickey
Great Questions and Posts! Could there be drag from the coriolis effect from rotational forces that might possibly slow down the oscillations as object fell back and forth through the hole in the earth? Even though the inward centripetal forces are greater at poles and outward centrifugal forces greatest at equator would there be any difference from a falling objects motion or a tendency to rotate in concert to the coriolis effect at poles or equator? I would think that a slight force would cause the object to rotate over time in alignment with the poles. Also, could Core Elements of earth provide a short period of greater gravity resistance as object passed through that zone causing object to possibly slow down over time and become equalized to the orbit of the core of the earth, or as you said maybe that could not happen because the greater mass would provide momentum to continue to oscillate?
John
John
Please Log in or Create an account to join the conversation.
- tvanflandern
- Topic Author
- Offline
- Platinum Member
Less
More
- Thank you received: 0
19 years 11 months ago #12326
by tvanflandern
Replied by tvanflandern on topic Reply from Tom Van Flandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by cosmicsurfer</i>
<br />Could there be drag from the coriolis effect from rotational forces that might possibly slow down the oscillations as object fell back and forth through the hole in the earth?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">True. The previous discussion was for a pole-to-pole hole through the Earth, for which rotation was not a factor.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Even though the inward centripetal forces are greater at poles and outward centrifugal forces greatest at equator would there be any difference from a falling objects motion or a tendency to rotate in concert to the coriolis effect at poles or equator?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">There is no coreolis force for an object falling vertically at the poles.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">could Core Elements of earth provide a short period of greater gravity resistance as object passed through that zone causing object to possibly slow down over time and become equalized to the orbit of the core of the earth, or as you said maybe that could not happen because the greater mass would provide momentum to continue to oscillate?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Gravity does not cause "resistance". It conserves momentum. But if you wanted to consider non-rigid bodies falling, then tidal forces might eventually slow down the oscillation. -|Tom|-
<br />Could there be drag from the coriolis effect from rotational forces that might possibly slow down the oscillations as object fell back and forth through the hole in the earth?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">True. The previous discussion was for a pole-to-pole hole through the Earth, for which rotation was not a factor.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Even though the inward centripetal forces are greater at poles and outward centrifugal forces greatest at equator would there be any difference from a falling objects motion or a tendency to rotate in concert to the coriolis effect at poles or equator?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">There is no coreolis force for an object falling vertically at the poles.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">could Core Elements of earth provide a short period of greater gravity resistance as object passed through that zone causing object to possibly slow down over time and become equalized to the orbit of the core of the earth, or as you said maybe that could not happen because the greater mass would provide momentum to continue to oscillate?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Gravity does not cause "resistance". It conserves momentum. But if you wanted to consider non-rigid bodies falling, then tidal forces might eventually slow down the oscillation. -|Tom|-
Please Log in or Create an account to join the conversation.
- rousejohnny
- Offline
- Elite Member
Less
More
- Thank you received: 0
19 years 11 months ago #12035
by rousejohnny
Replied by rousejohnny on topic Reply from Johnny Rouse
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by tvanflandern</i>
<br /><blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by rousejohnny</i>
<br />No, because drag from the centrifugal force is "slowing" them down. Is that even a good guess?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">At least you took a guess. Good effort. But centrifugal force is not a form of drag.
The answer is that gravitational potential is weaker on the equator than at the poles, which makes clocks at the equator run faster. The two effects, clock-slowing from speed and clock speed-up from potential, exactly cancel, leaving Earth's entire surface as an "equichron" (equal time) surface. -|Tom|-
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
What is it that makes the gravitational potential greater at the poles? Is it not the bulge of mass at the equator. Is this bulge not generated by the centrifugal force of a spinning sphere. Does the bulging not recend towards the poles using centrifugal considerations in determining the gravitational potential of a given latitude?
<br /><blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by rousejohnny</i>
<br />No, because drag from the centrifugal force is "slowing" them down. Is that even a good guess?<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">At least you took a guess. Good effort. But centrifugal force is not a form of drag.
The answer is that gravitational potential is weaker on the equator than at the poles, which makes clocks at the equator run faster. The two effects, clock-slowing from speed and clock speed-up from potential, exactly cancel, leaving Earth's entire surface as an "equichron" (equal time) surface. -|Tom|-
<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
What is it that makes the gravitational potential greater at the poles? Is it not the bulge of mass at the equator. Is this bulge not generated by the centrifugal force of a spinning sphere. Does the bulging not recend towards the poles using centrifugal considerations in determining the gravitational potential of a given latitude?
Please Log in or Create an account to join the conversation.
Time to create page: 0.317 seconds