# When will gravity lose its grip?

How much mass attraction is necessary for a stone (mass 1 kg) to rotate together with same velocity than the earth?

I mean in which altitude will the stone will no longer be able to follow the earth’s rotation.

Do it exist any equation to calculate this?

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The second the rock leaves the ground it is no longer following the earth's rotation.

The second the rock leaves the ground it is no longer following the earth's rotation.
So when jumping I am not following the Earths rotation ? Really?
I mean we ignore the atmosphere.

How can we calculate ?

Janus
Staff Emeritus
Gold Member
So when jumping I am not following the Earths rotation ? Really?
Really. The minute you leave the ground, you are following a path independent of the Earth's rotation. You actually follow an orbital path which intersects the surface of the Earth. Even if you jump straight up, you will not come back down exactly where you jumped. For low height jumps, you will not notice any difference, but as you increase the height of the jump the difference will increase.

How can we calculate ?
It takes some orbital mechanics. First you have to determine the properties of the orbital path you take from the initial speed of your jump and your linear velocity at your latitude due the Earth's rotation. Then you need to calculate the time it takes for you to travel the segment of your orbit from when you left the surface to when it intersects the Earth again. You will also need to calculate how many degrees of on orbit you traveled. Compare this to the number of degrees the Earth turned while you were "in the Air". Use the difference between the two and the radius of the Earth to find the distance between the point you jumped and the point where you landed.

DaveC426913
Gold Member
Really. The minute you leave the ground, you are following a path independent of the Earth's rotation.
But do not confuse this with following a path independent of Earth's gravitational pull.

DavidSnider
Gold Member
The earth is rotating at 1000 MPH Right? Obviously, when I jump the ground doesn't move 1000 MPH. I'm guessing that is because because my velocity when I jump is 1000 MPH too? If I were in a stationary position in space and looked at a jet travelling 1000MPH on earth would it appear to be fixed in place?

DaveC426913
Gold Member
If I were in a stationary position in space and looked at a jet travelling 1000MPH on earth would it appear to be fixed in place?
Assuming you qualify all those 1000mph's with 'at the equator', and
assuming the plane was headed East, and
assuming by 'fixed in place' you mean 'in your line of sight as the Earth turns under it', yes.

when you feet leave the ground you are veering on a path that is tangentional to the rotating path you were taking when your feet were on the ground which, yes, would be the same as the earth's rotation speed the moment before you jumped. Think of it like you're twirling a weight on a string above your head and then you let go. The thing flies of tangentionally at the speed your were spinning.

assuming by 'fixed in place' you mean 'in your line of sight as the Earth turns under it', yes.
well no, generally when we say that a plane is going 1000mph we're implying that we mean RELATIVE to the surface of the earth.

DaveC426913
Gold Member
well no, generally when we say that a plane is going 1000mph we're implying that we mean RELATIVE to the surface of the earth.

He is asking if the plane, travelling at 1000mph would remain 'fixed in place' to an observer stationary in space. I am simply pointing out that 'fixed in place' is ambiguous. 'Fixed in place' would normally mean relative to the Earth. In this case, I'm hoping the poster means 'fixed in your line of sight'.

russ_watters
Mentor
when you feet leave the ground you are veering on a path that is tangentional to the rotating path you were taking when your feet were on the ground....
Not unless you jump straight sideways you aren't!

When you are standing still on the ground, your direction of motion is tangential to the rotating path of the surface of the earth. When you jump, your direction of motion is some angle above the tangent. Not a big angle, but an angle, nonetheless.

Not unless you jump straight sideways you aren't!

When you are standing still on the ground, your direction of motion is tangential to the rotating path of the surface of the earth. When you jump, your direction of motion is some angle above the tangent. Not a big angle, but an angle, nonetheless.
Sorry, my bad. The TANGENTIONAL component, or the component with a projection on the rotation axis of the earth, will be the velocity of the earth's rotation. Your full velocity will include the small vertical jumping velocity as well.

rcgldr
Homework Helper
I mean in which altitude will the stone will no longer be able to follow the earth’s rotation.
Without the atmoshpere, the stone traveling east at the equator would orbit the earth faster than the earth rotates until it's at an altitude of 35,786 km == 22,236 miles above ground, where the orbital rate would match up with the surface of the earth:

http://en.wikipedia.org/wiki/Geostationary_orbit

At higher altitudes, it would orbit slower.

Which law of nature is responsible for that the atmosphere follows the Earth's surface?
Density off course, - but which equation?

DaveC426913
Gold Member
Which law of nature is responsible for that the atmosphere follows the Earth's surface?
Density off course, - but which equation?
What?

The atmosphere is held to the Earth by gravity. It forms a sphere whose shape does not reflect the underlying altitude of the terrain. Just ask the people in Denver Colorado, who live at 5000ft.

russ_watters
Mentor
I think he's asking why the atmosphere rotates with the earth. There are two ways to look at it, both using Newton's First law:
1. There is no force that would make the atmosphere stop rotating.
2. If the atmosphere were not rotating with earth, there'd be aerodynamic drag causing it to speed up and rotate with the earth.

I think he's asking why the atmosphere rotates with the earth.[quote fixed]
Right...

If the atmosphere were not rotating with earth, there'd be aerodynamic drag causing it to speed up and rotate with the earth.
What do you mean: "aerodynamic drag".

Lets image we could stop the atmosphere rotating, - and push the STOP.
How fast would it begin to rotate again after pushing START ?

Does any equation exist to calculate this?
I mean what happens in higher altitude..

russ_watters
Mentor
What do you mean: "aerodynamic drag".
I mean aerodynamic drag! It's the thing that pushes your hand back when you stick it out the window of a moving car.
Lets image we could stop the atmosphere rotating, - and push the STOP.
How fast would it begin to rotate again after pushing START ?
If you could "stop" the atmosphere rotating, then the wind speed at ground level near the equator would be about 1000 mph. It would rip anything and everything off the ground, but in so doing, those things would slow the wind down a little. And even after the earth was swept mostly smooth in those first few seconds, there'd be friction between the wind and the ground. A lot of friction. It would take only minutes for the wind at ground level to lose most of that speed. Then viscous friction between molecules in the air would eventually slow the rest of the atmosphere. Would it take days or weeks? I don't know. But it wouldn't be a very long time. It certainly wouldn't be years.

Really. The minute you leave the ground, you are following a path independent of the Earth's rotation. You actually follow an orbital path which intersects the surface of the Earth. Even if you jump straight up, you will not come back down exactly where you jumped. For low height jumps, you will not notice any difference, but as you increase the height of the jump the difference will increase.
so what is the difference between earth's rotation and an orbital path?

It takes some orbital mechanics. First you have to determine the properties of the orbital path you take from the initial speed of your jump and your linear velocity at your latitude due the Earth's rotation. Then you need to calculate the time it takes for you to travel the segment of your orbit from when you left the surface to when it intersects the Earth again. You will also need to calculate how many degrees of on orbit you traveled. Compare this to the number of degrees the Earth turned while you were "in the Air". Use the difference between the two and the radius of the Earth to find the distance between the point you jumped and the point where you landed.
I didn't quite understand what you meant by the orbit intersecting with the Earth

Janus
Staff Emeritus
Gold Member
so what is the difference between earth's rotation and an orbital path?
When you are standing on the surface of the Earth and moving with its rotation you are not in free fall, when you are in an orbital path you are.

Janus
Staff Emeritus
Gold Member
I didn't quite understand what you meant by the orbit intersecting with the Earth
In the following image the green circle is the surface of the Earth, the red dot its center, and the black ellipse the orbital path. The Earth is rotating clockwise.
As you jump up, you are moving up and to the right with with the Earth. This puts you in a path that is an ellipse with one focus at the center of the Earth. We are concerned with the part from when you leave the surface to when you return to the surface. This is the part of the ellipse outside of the green circle.

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In the following image the green circle is the surface of the Earth, the red dot its center, and the black ellipse the orbital path. The Earth is rotating clockwise.
As you jump up, you are moving up and to the right with with the Earth. This puts you in a path that is an ellipse with one focus at the center of the Earth. We are concerned with the part from when you leave the surface to when you return to the surface. This is the part of the ellipse outside of the green circle.

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I didn't quite understand the diagram. When you jump shouldn't you have the same linear velocity as the earth so you would end up landing at the same spot but in that diagram it is more like a projectile motion where the object travelled faster than the earth's rotation. Also I thought that the center of ellipses were at the center of the Earth but that does not seem to be the case in that diagram. I think i'm misunderstanding the diagram somehow, can you clarify some of my misconceptions? Thanks for any help that you can provide

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Janus
Staff Emeritus
Gold Member

I didn't quite understand the diagram. When you jump shouldn't you have the same linear velocity as the earth so you would end up landing at the same spot but in that diagram it is more like a projectile motion where the object travelled faster than the earth's rotation. Also I thought that the center of ellipses were at the center of the Earth but that does not seem to be the case in that diagram. I think i'm misunderstanding the diagram somehow, can you clarify some of my misconceptions? Thanks for any help that you can provide
I modified the diagram to illustrate the point better. The diagram shows the view from someone not sharing the Earth's rotation. The angle between the red lines is how much the Earth rotates while you are in the air. Note that it rotates a bit more than the arc you travel from jumping to landing.

As to the ellipse, you are mistaken. The center of the Earth would be located at one of the foci of the ellipse, not its center. It is an orbital trajectory, and this is a property of elliptical orbits.

Here's a little on elliptical orbits:

I modified the diagram to illustrate the point better. The diagram shows the view from someone not sharing the Earth's rotation. The angle between the red lines is how much the Earth rotates while you are in the air. Note that it rotates a bit more than the arc you travel from jumping to landing.

As to the ellipse, you are mistaken. The center of the Earth would be located at one of the foci of the ellipse, not its center. It is an orbital trajectory, and this is a property of elliptical orbits.

Here's a little on elliptical orbits: