Why aren't we heavier during the day?

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  • #1
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Ok, seems like a silly question, but this one has been puzzling me. I'm sure there is an easy answer, but it is not an easy thing to google.

So imagine swinging a bucket of water round. The water can be inverted, but it stays in the bucket with centrifugal force. And centrifugal force occurs whenever something changes direction. So my question is, as earth is orbiting the sun, and hence changing direction, why do we not feel a centrifugal force? IE be pushed into the earth during the day, and be flung outwards (become lighter) during the night. I was thinking relativity, but gyros maintain their direction relative to space on earth regardless of spinning. So why don't we feel centrifugal force as the earth races round the sun?
 

Answers and Replies

  • #2
I believe I understood the question and my only possible answer would be that our center of gravity towards the earth does not change, so unless you move further away from the earths center (core) (theoretically) you will not lose weight.
 
  • #3
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Thank you. My question is more based on location upon the surface. Why during the day when centrifugal force would throw you into the earth, are you not lighter than at night, when on the opposite side of earth, and centrifugal force would be trying to throw you away from the surface?
 
  • #4
Ahh, I see. I believe you're right we do weigh less but by such a minuscule amount that you would probably NEVER know your own weight changed.

This is what my physics teacher has helped me to understand at least, I hope this cleared up anything if not, I wish you the best of luck in your search of knowledge.
 
  • #5
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Ahh, I see. I believe you're right we do weigh less but by such a minuscule amount that you would probably NEVER know your own weight changed.

This is what my physics teacher has helped me to understand at least, I hope this cleared up anything if not, I wish you the best of luck in your search of knowledge.
Thanks, I think I am not grasping the physics of the situation.
Centrifugal force is F=(m*v^2)/r

Assuming earth goes round the sun at 30,000 m/s
Distance to sun (not centre of but ball park) is 150,000,000,000 m

So for a man with a *mass of 70 kilos I get roughly

70*(30000^2)/150,000,000,000

= 0.42Newtons

Well that's quite a lot. I mean that being so, not only should space launches be done at the equator, it would very much pay to do them at night as well. And it is a diurnal difference of 1 kg in just my weight.
 
  • #6
Thanks, I think I am not grasping the physics of the situation.
Centrifugal force is F=(m*v^2)/r

Assuming earth goes round the sun at 30,000 m/s
Distance to sun (not centre of but ball park) is 150,000,000,000 m

So for a man with a *mass of 70 kilos I get roughly

70*(30000^2)/150,000,000,000

= 0.42Newtons

Well that's quite a lot. I mean that being so, not only should space launches be done at the equator, it would very much pay to do them at night as well. And it is a diurnal difference of 1 kg in just my weight.
I'm not entirely sure but don't you have to convert the meters to KM?
EDIT: Also, why are you using the distance to the sun as your radius?
 
  • #8
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Also, why are you using the distance to the sun as your radius?
Laziness. It should be distance to the sun + the sun's radius to give our radius about the orbit, but it was close enough for a ball park figure. I'm not going to be out by a massive factor, but saved my 2 mins research. :)
 
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  • #9
= 0.42Newtons...Well that's quite a lot.
I also calculate less than 1 Newton, which is roughly the weight of an apple, less than 0.5% of our body weight. Our own digestion and sweating causes more weight change than this is a day and no one really notices. Legs are quite strong and efficient at their job.
 
  • #10
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I also calculate less than 1 Newton, which is roughly the weight of an apple, less than 0.5% of our body weight. Our own digestion and sweating causes more weight change than this is a day and no one really notices. Legs are quite strong and efficient at their job.
What, so we get heavier? Space launches are done at the equator for a saving of 0.5% of the gravity between pole and equator. Probably less than if they picked an arbitrary point like New York that had a mid latitude position. But they can save as much again by launching at night? I'd be in line for a nobel prize for saving them all that money.

I'm sure I have messed up the physics of the situation. I have not grasped something, be it relativity or something else.
 
  • #11
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Lol you are all noobs, the "centrifugal force" as you call it, is caused by gravitational force. The sun is pulling you directly (because you have mass). It's not like the bucket and water where the bucket is pushing the water.

The earth is not tied with a rope and being swung around.
 
  • #12
cjl
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There's no difference, because the sun's gravity directly affects you. If you were unaffected by the sun's gravity, then the calculation would be correct. However, the sun's gravity does affect you. If the earth vanished, you would still be in orbit around the sun, so no force is required from the earth to maintain this curvature.
 
  • #13
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I have seen my mistake
 
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  • #14
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Same thing with freefall. You are accelerating at ~10m/s^2 but you feel nothing because each particle in your body is experiencing uniform force.

If you are in a car accelerating at 10m/s^2 its a different story because the chair is pushing you.
 
  • #15
russ_watters
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What, so we get heavier? Space launches are done at the equator for a saving of 0.5% of the gravity between pole and equator. Probably less than if they picked an arbitrary point like New York that had a mid latitude position. But they can save as much again by launching at night? I'd be in line for a nobel prize for saving them all that money.

I'm sure I have messed up the physics of the situation. I have not grasped something, be it relativity or something else.
Space launches are done near the equator to gain the speed of the earth's rotation.
 
  • #16
D H
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There's no difference, because the sun's gravity directly affects you. If you were unaffected by the sun's gravity, then the calculation would be correct. However, the sun's gravity does affect you. If the earth vanished, you would still be in orbit around the sun, so no force is required from the earth to maintain this curvature.
There is a difference. The vector from the center of the Sun to you with have a different magnitude and/or direction than the vector from the center of the Sun to the center of the Earth. This results in sunward gravitational acceleration vectors that differ by a small amount. That these acceleration vectors are not equal is what that causes the tides, the solar tides in this case.

The difference is quite small. It points away from the Earth at noon and at midnight, making you slightly lighter. It points inward at sunrise and sunset, but with half the magnitude of the outward tidal force at noon and midnight.
 
  • #17
Thanks, I think I am not grasping the physics of the situation.
Centrifugal force is F=(m*v^2)/r

Assuming earth goes round the sun at 30,000 m/s
Distance to sun (not centre of but ball park) is 150,000,000,000 m

So for a man with a *mass of 70 kilos I get roughly

70*(30000^2)/150,000,000,000

= 0.42Newtons

Well that's quite a lot. I mean that being so, not only should space launches be done at the equator, it would very much pay to do them at night as well. And it is a diurnal difference of 1 kg in just my weight.
Hi,
0.42 newtons would be mean force, as you took mean distance. If you want to get difference between night and day you should take

F_day=(m*v^2)/(distance_to_sun - earth_radius)
F_night=(m*v^2)/(distance_to_sun + earth_radius)

your GAIN or LOSS of weight would be F_gain=(F_day - F_night), which would be much much less than 0.42N, say (exaggerating) 0.421N during the day and 0.419N during the night gives you only 0.002N difference. In reality this number would be smaller of course.
 
  • #18
Redbelly98
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So for a man with a *mass of 70 kilos I get roughly

70*(30000^2)/150,000,000,000

= 0.42Newtons

Well that's quite a lot.
0.4 Newtons is only 0.04 kg. That's the mass of a light snack, and just 0.06% of the total 70 kg mass.
 
  • #19
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Also 0.4 Newtons is like 0.08 lbs!

But hey, you should try an experiment. Get a scale that you can plug in to your computer, set up a script to take a measurement of some mass every hour or something and then plot the results by time of day.
 
  • #20
cjl
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There is a difference. The vector from the center of the Sun to you with have a different magnitude and/or direction than the vector from the center of the Sun to the center of the Earth. This results in sunward gravitational acceleration vectors that differ by a small amount. That these acceleration vectors are not equal is what that causes the tides, the solar tides in this case.

The difference is quite small. It points away from the Earth at noon and at midnight, making you slightly lighter. It points inward at sunrise and sunset, but with half the magnitude of the outward tidal force at noon and midnight.
True, there would be tidal forces, but they would be of a much smaller magnitude than the effect that the OP was discussing. I doubt they'd be measurable without fairly expensive equipment.
 
  • #21
D H
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Hi,
0.42 newtons would be mean force, as you took mean distance. If you want to get difference between night and day you should take

F_day=(m*v^2)/(distance_to_sun - earth_radius)
F_night=(m*v^2)/(distance_to_sun + earth_radius)[

your GAIN or LOSS of weight would be F_gain=(F_day - F_night), which would be much much less than 0.42N, say (exaggerating) 0.421N during the day and 0.419N during the night gives you only 0.002N difference. In reality this number would be smaller of course.
In reality the number would be orders of magnitude smaller. Read [post=2968343]post #16[/post]. What would be seen for a person whose mass is 70 kg would be 35 micronewtons lighter at noon and midnight, 18 micronewtons heavier at sunrise and sunset.
 

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