Rotation effect on gravity?

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I read that Earth would have to spin at 28,437 km per hour to cause us to lift off the surface. I’m assuming at that point we would  effectively be weightless.  Seemingly throwing a wrench into F=GM1M2/R , so what effect would it have on the moon?

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6 minutes ago, jajrussel said:

what effect would it have on the moon?

In Newtonian physics, none.

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5 minutes ago, jajrussel said:

I read that Earth would have to spin at 28,437 km per hour to cause us to lift off the surface. I’m assuming at that point we would  effectively be weightless.  Seemingly throwing a wrench into F=GM1M2/R ,

Why do you think it would "throw a wrench into into F=GM1M2/R ?

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14 minutes ago, jajrussel said:

I read that Earth would have to spin at 28,437 km per hour to cause us to lift off the surface. I’m assuming at that point we would  effectively be weightless.  Seemingly throwing a wrench into F=GM1M2/R , so what effect would it have on the moon?

There is a speed such that the centripetal force is equal to the gravitational force. v = sqrt(GM/r).

That’s the speed of a circular orbit at r. You would be weightless, but not lift up. (but the earth would fall apart before this could happen)

Far from throwing a wrench into it - the above equation uses the equation to solve for v

The mass of the earth would be slightly larger, increasing the pull on the moon by a small amount. Any other effect on the moon would be found in GR

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Just now, Bufofrog said:

Why do you think it would "throw a wrench into into F=GM1M2/R ?

Doesn’t F=GM1M2/Rapply to me and the earth? At a spin of 28,437 km per hour we lift off. My assumption is that in order for that to happen the acceleration due to spin would have to exceed the forces connecting us.

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3 minutes ago, jajrussel said:

Doesn’t F=GM1M2/Rapply to me and the earth? At a spin of 28,437 km per hour we lift off. My assumption is that in order for that to happen the acceleration due to spin would have to exceed the forces connecting us.

There’s no lift - there’s no upward force

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4 minutes ago, swansont said:

There is a speed such that the centripetal force is equal to the gravitational force. v = sqrt(GM/r).

That’s the speed of a circular orbit at r. You would be weightless, but not lift up. (but the earth would fall apart before this could happen)

Far from throwing a wrench into it - the above equation uses the equation to solve for v

The mass of the earth would be slightly larger, increasing the pull on the moon by a small amount. Any other effect on the moon would be found in GR

Thanks , I don’t write or think fast enough. I’ll read and think on this a while.

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15 minutes ago, jajrussel said:

I read that Earth would have to spin at 28,437 km per hour to cause us to lift off the surface. I’m assuming at that point we would  effectively be weightless.  Seemingly throwing a wrench into F=GM1M2/R , so what effect would it have on the moon?

1.  It is meaningless to say "spin at 28,437 km", as rotation needs to be measured as angular velocity. (deg/hr, rad/sec. etc.)

I know that it is common to express this in terms of tangential speed( in this case, at the equator) But it is sloppy and can lead to misunderstanding. For example if the Earth had a tangential speed of 28,437 kph at the equator, then at 45 north latitude it would only be 20,105 kph

As swansont has pointed out there is a speed where the centripetal force (the force required to keep an object moving in circle) and the gravitational force balance out. This would result in the object going into orbit around the Earth.   Gravity is still in play. In fact it is gravity that would prevent someone standing on the equator from just shooting out into space at in a straight line instead of just hovering over the Earth. And by the way, your number is a bit low, the equatorial speed would need to be 28463 kph

And because of what I alluded to earlier, only someone on the equator would even go into orbit.  People elsewhere will feel lighter, and the ground would seem to tilt a bit under their feet ( And even this is an over simplification which assumes the Earth maintains its present shape.  If the Earth was indeed spinning this fast, its very shape would change, making it much more of an oblate spheroid.

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1 hour ago, swansont said:

There’s no lift - there’s no upward force

This is where I got the word lift. Not arguing just saying. Not certain if I added this picture correctly? Guess I have to hit submit to find out… note it is not a link, just part of a screenshot.

Edited by jajrussel
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It's just sloppy language. From the point of view of a person anchored to the ground, another person might appear to lift in this scenario.

But it's not lift, it's more that the ground is "falling away from" the un-anchored person, and they continue to move as they did.

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24 minutes ago, jajrussel said:

This is where I got the word lift. Not arguing just saying. Not certain if I added this picture correctly? Guess I have to hit submit to find out… note it is not a link, just part of a screenshot.

As I already noted, 28,437 kph falls a bit below that for even a near surface orbit.   To be lifted off "into space", using the Kármán line (at 100 km altitude) for the boundary of space, you would need to be moving at 28,498.5 kph.  In which case, you would rise to a height of 100 km (the apogee of your orbit), and then drop back down to perigee at the surface, rise up to apogee...

To leave the Earth's vicinity entirely, you'd need to be moving at 40,253 kph ( escape velocity)

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The gravitational acceleration on the Earth surface is g=9.8 m/s2. A body moving with the speed v=28,437 km/h and accelerating toward a center with the g would rotate at the distance R=v2/g=63,670 km from the center. This distance, 63,700 km, is about the Earth radius. Thus, this body would rotate around the Earth center without pushing on the ground, "levitating".

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Wow, Earth has really gotten fat!

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4 minutes ago, TheVat said:

Wow, Earth has really gotten fat!

Evidently, "Details matter" in the other thread...

Let's fix it:

The gravitational acceleration on the Earth surface is g=9.8 m/s2. A body moving with the speed v=28,437 km/h and accelerating toward a center with the g would rotate at the distance R=v2/g=6,367 km from the center. This distance, 6,370 km, is about the Earth radius. Thus, this body would rotate around the Earth center without pushing on the ground, "levitating".

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15 hours ago, TheVat said:

Wow, Earth has really gotten fat!

Be nice! It's still a heavenly body...

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