Force needed to lift the Earth?

[The Grim Squeaker]

Having recently looked at the famous tale of Hercules holding up the planet, I was curious as to how muc force he would need to exert for this feat.

The mass of the Earth = 5.9742×10^24 kilograms [according to a quick Google].

Now, the question is how much gravity would be exerted, by the sun and other planets as well as the famous gravitational constant.
If the Gravity is merely Earth's normal exerted gravity or that exerted by the sun then wouldn't the Force needed be F=ma with a=the gravity and m=mass of the earth (Or the "Weight" of the earth affected by the gravity of the sun?).

For the sake of the argument the lifter can stand in air and can extend a field so that the force exerted on the planet is spread out and doesn't rip the atmosphere/oceans/etc' away.

[Darth Wong]

Do a handstand. There; you just exerted enough force to lift the Earth ... relative to your own torso. Or did you forget that you have to define an external frame of reference for this?

[The Grim Squeaker]

Do a handstand. There; you just exerted enough force to lift the Earth ... relative to your own torso. Or did you forget that you have to define an external frame of reference for this?

Heh, good point. (Sorry for being Vague).

I mean the force necessary to either stop the earth in it's orbit, to push it out of it's orbit or to "hold" it in place while preventing it from moving.

If these examples are too obscure then the force needed to Hold the Earth and lift it and hold it in place is also useful (Similair to the famous Hercules/Atlas feat of myth), the question is how much gravitational force would be exerted by the sun and other objects in the solar system on it.

[Prozac the Robert]

I mean the force necessary to either stop the earth in it's orbit, to push it out of it's orbit or to "hold" it in place while preventing it from moving.

I supose you could work out the force required to hold the earth still relative to the sun.

You'd need to do work to stop the circular motion, and then a constant force of GMm/(r^2) to hold it against the sun's gravity.

Rotation:
The Earth apparently has mean orbital velocity of 29.78 km/s, or 2.978*10^2 m/s.
Therefore it has a kinetic energy of 2.965*10^23 Joules. That's how much workwould need to be done to stop that. The force required depends on how long you have to do that.

Radial force:
This requires a constant force of 3.56*10^22 Newtons, provided I have that calculated right.

[AK_Jedi]

The sun exerts approx. 3.5x10^22 newtons of force on the Earth. If the Earth were to stop revolving around the sun, this is how much force you would need to hold it up, not taking the moon into account. Most of the time, the other planets exert negligible force on the Earth.

[Dooey Jo]

Rotation:
The Earth apparently has mean orbital velocity of 29.78 km/s, or 2.978*10^2 m/s.
Therefore it has a kinetic energy of 2.965*10^23 Joules.

29.78 km/s is actually 2.978e4 m/s...
Also, the mass of the Earth is about 6e24 kg, so the kinetic energy,
E = (m*v^2) / 2
would be
(6e24 * (2.978e4)^2) ) / 2 = ~3e33 J

[Prozac the Robert]

29.78 km/s is actually 2.978e4 m/s...
Also, the mass of the Earth is about 6e24 kg, so the kinetic energy,
E = (m*v^2) / 2
would be
(6e24 * (2.978e4)^2) ) / 2 = ~3e33 J

Umm, no idea why I wrote that speed. I plead temporary stupidity.

Also, that speed doesn't seem to give the answer I gave anyway, so at least it's possible that I didn't type 2.978e2 into the calculator. I think I must have just mashed some wrong buttons. I've just finished the third year of a physics degree, and I can't even correctly operate a calculator, or reliably write down a number in standard form. How pathetic is that.

Thanks for sorting that out.

[Molyneux]

Having recently looked at the famous tale of Hercules holding up the planet, I was curious as to how muc force he would need to exert for this feat.

The mass of the Earth = 5.9742×10^24 kilograms [according to a quick Google].

Now, the question is how much gravity would be exerted, by the sun and other planets as well as the famous gravitational constant.
If the Gravity is merely Earth's normal exerted gravity or that exerted by the sun then wouldn't the Force needed be F=ma with a=the gravity and m=mass of the earth (Or the "Weight" of the earth affected by the gravity of the sun?).

For the sake of the argument the lifter can stand in air and can extend a field so that the force exerted on the planet is spread out and doesn't rip the atmosphere/oceans/etc' away.

Just a nitpick..but Hercules did NOT hold up the Earth. He was, in fact, standing *on* the Earth and holding up the sky, taking the job over from Atlas temporarily.

[Qwerty 42]

Having recently looked at the famous tale of Hercules holding up the planet, I was curious as to how muc force he would need to exert for this feat.

The mass of the Earth = 5.9742×10^24 kilograms [according to a quick Google].

Now, the question is how much gravity would be exerted, by the sun and other planets as well as the famous gravitational constant.
If the Gravity is merely Earth's normal exerted gravity or that exerted by the sun then wouldn't the Force needed be F=ma with a=the gravity and m=mass of the earth (Or the "Weight" of the earth affected by the gravity of the sun?).

For the sake of the argument the lifter can stand in air and can extend a field so that the force exerted on the planet is spread out and doesn't rip the atmosphere/oceans/etc' away.

Just a nitpick..but Hercules did NOT hold up the Earth. He was, in fact, standing *on* the Earth and holding up the sky, taking the job over from Atlas temporarily.

Although the mental image of Atlas standing on his hands for years at a stretch is thoroughly amusing.

[The Grim Squeaker]

Having recently looked at the famous tale of Hercules holding up the planet, I was curious as to how muc force he would need to exert for this feat.

The mass of the Earth = 5.9742×10^24 kilograms [according to a quick Google].

Now, the question is how much gravity would be exerted, by the sun and other planets as well as the famous gravitational constant.
If the Gravity is merely Earth's normal exerted gravity or that exerted by the sun then wouldn't the Force needed be F=ma with a=the gravity and m=mass of the earth (Or the "Weight" of the earth affected by the gravity of the sun?).

For the sake of the argument the lifter can stand in air and can extend a field so that the force exerted on the planet is spread out and doesn't rip the atmosphere/oceans/etc' away.

Just a nitpick..but Hercules did NOT hold up the Earth. He was, in fact, standing *on* the Earth and holding up the sky, taking the job over from Atlas temporarily.

Although the mental image of Atlas standing on his hands for years at a stretch is thoroughly amusing.

True.
Man I feel Humiliated, I who had read the Odyssey (Translated heavily albeit and in a illustrated formal) At age 8 misinterpreting this.
The shame I would request a boon to Kill myself but Seppuku is Japanese so scratch that .
Still The weight os the sky according to Here is 5.3 x 10^18 kg.
So Lifting it and holding it would require overcoming Earth's gravitational pull (Standardized, though it would weaken due to the ranges involved):
F=ma
m=5.3 x 10^18 kg
a=g=9.81 m/s^2
So Herc would need to sustain a force of 5.1993 × 10^19 Newtons.
We know that he needed to switch with Atlas and didn't lift it any further so this is the limit and not a low end limit (As would be the case for lifting the Sky up).

Also, the mass of the Earth is about 6e24 kg, so the kinetic energy,
E = (m*v^2) / 2
would be
(6e24 * (2.978e4)^2) ) / 2 = ~3e33 J

Heh, who needs the DS (Yeah, yeah 1e38...).
Thanks, I didn't know the speed involved in the earths rotation and whether just the rotation speed needed to be taken into account (Gravitational pull from other Celestial objects etc' spring to mind).
Thanks for the answers