Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: The Scientist on 09/07/2010 09:45:23
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How many kilometers would you have to go above the surface of the earth for your weight to decrease to half of what it was at the surface?
Please explain your answer. Thank you!
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According to Newton 2641.9 km
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Remember though you would need a solid surface fixed to the earth to experience "weight" the inverse square law is OK for the poled because there is no net rotation however if your tower was on the equator it could be slightly lower because of the increased centripetal force. Remember also that as there is no solid surface to stand on at that height to stay at that height you would have to be in orbit around the earth and the centripetal force of the orbital velocity would precisely cancel out the effect of "weight" and you would be weightless
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Weight a minute!
You wouldn't have to be orbiting the Earth if you were standing on the tower, so you would weigh something (unless that happened to put you in a geostationary orbit).
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According to Newton 2641.9 km
Can you explain how you calculated that please?
Chris
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According to Newton the force of gravity falls off as the square the distance of the bodies that are attracted to each other.
Your origanal distance from the centre of the Earth when you are on the surface is 6378.1 Km, if you wish to reduce the attraction to half what it is at the surface you must increase this distance to square root 2 times what it was hence you must be 2641.9 Km above the surface.
This attractive force will only be apparent as weight if you are supported on a solid surface if you were orbiting in a satellite you would have no sensation of weight.
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if you were orbiting in a satelite you would have no sensation of weight.
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Just a comparison not to divert from the original question.
Is being in orbit essentially the same as being on a parabolic ride inside a 747 Jumbo jet?
The experience on the testing or human conditioning for weightlessness in free air.
Aren't both situations considered as a free fall which sets the conditions that cancels the effects experienced from gravity?
Even a geostationary orbit is still an orbit that experiences the same affects, except the same ground is always in view because of the the orbit is in the same direction and has the same speed of the earth's rotation, which is aways matching in that particular case?
If this is so, then the only way is to calculate the distance altitude where gravity disipates by how much.
How can you have a direct scale measurement where only gravity to ground reference is the only two components in the affect of measurement, highest ground point is Mount Everest, realistically speaking that is.
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According to Newton the force of gravity falls off as the square the distance of the bodies that are attracted to each other.
Your origanal distance from the centre of the Earth when you are on the surface is 6378.1 Km, if you wish to reduce the attraction to half what it is at the surface you must increase this distance to square root 2 times what it was hence you must be 2641.9 Km above the surface.
This attractive force will only be apparent as weight if you are supported on a solid surface if you were orbiting in a satelite you would have no sensation of weight.
Never thought of it in this way...Thanks...
Gravity decreases with altitude, since greater altitude means greater distance from the Earth's centre. All other things being equal, an increase in altitude from sea level to the top of Mount Everest (8,850 metres)causes a weight decrease of about 0.28%.
Solve for h; set gh = .5 go and plug and chug
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Where:
gh is the gravity measure at height above sea level.
re is the Earth's mean radius.
go is the standard gravity
http://en.wikipedia.org/wiki/Gravity_of_Earth
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Tommya300
The path in which a body must move above the Earth to experience 'weightlessness' is elliptical not parabolic, in the case of an aircraft the difference is small but if you sent your satellite on a parabolic path it would disappear into space.
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Tommya300
The path in which a body must move above the Earth to experience 'weightlessness' is elliptical not parabolic, in the case of an aircraft the difference is small but if you sent your satellite on a parabolic path it would disappear into space.
Thank for that point.
Your right there what was I thinking. Maybe with not enough energy to send the satellite on a permanent parabolic journey it just might take an elliptical course?
All I wanted to know if an orbit is a free fall where that affect cancels the affect of gravity.
Can you help me with solving this equation for h
would gh = .5 go to get to .5 of the original weight at sea level?
I am having a small brain fart from time to time.
I solved for h= (1.414 re)- re
I keep on coming up with 9012255.3 meters from the earth's center
2634155.3 meters from sea level
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Syhprum, surely any path that is followed in which the aircraft is in freefall will allow the occupants to feel weightless, to be continuous you are right it must be an orbit. the vomit comet (which i think tommy was referring to) is described as making parabolas (http://jsc-aircraft-ops.jsc.nasa.gov/Reduced_Gravity/about.html) - whether these are in reality sections of an ellipse I don't know. A simple object projected skywards at an initial speed will describe a parabola, but how friction, the aerodynamic surfaces and huge speeds/distances involved will change this is beyond my ken. Matthew
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So assuming that you would always weigh something if standing on top of a tall tower, what would you weight at 35,786 km - the height at which Geosynchronous satellites orbit? I see a contradiction here...
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The usual calculation for a projectile assumes that gravity always acts "downwards" and this is near enough to true for rifle shells and high jumpers.
http://en.wikipedia.org/wiki/Trajectory
It's not correct because the attraction is towards a point rather than a plane. Once you talk about really fast things and large distances you need to use the correct form of the equations and you get an ellipse.
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So assuming that you would always weigh something if standing on top of a tall tower, what would you weight at 35,786 km - the height at which Geosynchronous satellites orbit? I see a contradiction here...
That does not seem too inconsistent to me. I think that says that when the tower is 35,786 km high, you weigh nothing.
Does that mean that when the tower is higher than that, you have negative weight?
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Thanks Bored Chemist all this talk of 'parabolas' always annoys me !
Fozzie
When you reach a height above the surface of the Earth of 35786 Km (6.611 times further from the centre of the Earth) gravity will be reduced to 2.289 % of its surface value.
This would only manifest itself as weight if you were standing on a tower above one of the poles.
If you were orbiting the equator on a 24 hr orbit as geostationary satellites do the centripetal force would cancel this out and you would be weightless.
Geezer Yes you would float away if the top of the tower was moving at geosynchronous orbit speed!
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Tommya300
The reason for the small error in your calculation is because you take 2^.5 to be 1.414, this value is too small if you take a more nearly correct value 1.4142136 I think it will remove the error.
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BC I thought he was using Geosynchronous Satellite orbit just to establish he is looking at a fixed position with respect to ground a weighing ground point. Hypothetically looking at just the earth's gravity and no other forces what will the gravitational different at the altitude of 35786 km?
Geosynchronous orbit is described as a direct, circular, low-inclination orbit around Earth having a period of 23 hours 56 minutes 4 seconds and a corresponding altitude of 35,784 km (22,240 miles, or 6.6 Earth radii).
Te affective acceleration gravity with no other affects, at Geosynchronous altitude of 35784 km,
I got 1.9116132455291657563746449861115e-13 m/s sq.
1.9506257607440466901782091695016e-12 % of the weight at sea level
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Tommya300
The reason for the small error in your calculation is because you take 2^.5 to be 1.414, this value is too small if you take a more nearly correct value 1.4142136 I think it will remove the error.
OKAAAA I am in a ballpark after all my rounding out. Thanks
actually I did 1/(.5^.5) and recipricated that again
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So assuming that you would always weigh something if standing on top of a tall tower, what would you weight at 35,786 km - the height at which Geosynchronous satellites orbit? I see a contradiction here...
That does not seem too inconsistent to me. I think that says that when the tower is 35,786 km high, you weigh nothing.
Does that mean that when the tower is higher than that, you have negative weight?
I think it may mean the weight is that much closer to a rediculously small number
BTW since there was no baseline set to the weight at sea level a number only can refer to it as in percentage
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Geezer Yes you would float away if the top of the tower was moving at geosynchronous orbit speed!
Well, it would kind of have to be, or it wouldn't be a very good tower, would it [;D]
Mind you, I suppose we are neglecting the actual mass of the tower here, but why would we want to mess up a perfectly good thought experiment with minor practicalities like that?