Understanding Gravitational Force: Why Does My Effect on Earth Seem Negligible?

[cosmogrl]

OK, so the force between me and Earth is G*m1*m2/r^2, and I have the same pull on Earth as Earth has on me. I see it in the math, BUT, it doesn't make sense to me outside of the math. Earth's gravity pulls me down to the surface...certainly I don't pull up on Earth with the same force? Why is my effect on Earth negligible when compared to Earth's effect on me? Looking at the equation above I would think my effect on Earth is the same as Earth's effect on me. Can you help me understand this? Thanks.

 

[Chrisas]

...because of F=ma. The law of action and reaction requires the force the Earth acts on you to be equal in magnitude to the force you act on the earth. However, when you put the equation for Fg on one side of F=ma, then look at the acceleration effect.

You accelerate due to the earth:
G*m_you*m_earth/Re^2 = m_you * a_you
The small m_you cancels and the a_you depends on the mass of the Earth only. You should get something like 9.8 m/s/s or 32 ft/s/s when you plug in Earth parameters.

Earth accelerates due to you:
G*m_you*m_earth/Re^2 = m_earth *a_earth

The ginormous m_earth cancels and leaves a_earth dependent on the very small m_you divided by the rather large Re^2.

 

[dave_baksh]

As above but in layman's, the forces are the same. However, it takes more force to move the Earth than it does to move you, so it seems like you're being pulled to Earth and not being pulled to your common centre of gravity.

For funsies, imagine jumping on the spot once. You might think you go up and then come back down but in reality when you jump, you and the Earth are moving in opposite directions, until the force of mutual attraction becomes dominant and you're sucked back together.

 

[Klockan3]

According to GR when you jump you are not accelerating downwards towards the earth, it is instead the Earth's surface which is accelerating upwards and catching you mid air.

 

[ZapperZ]

OK, so the force between me and Earth is G*m1*m2/r^2, and I have the same pull on Earth as Earth has on me. I see it in the math, BUT, it doesn't make sense to me outside of the math. Earth's gravity pulls me down to the surface...certainly I don't pull up on Earth with the same force? Why is my effect on Earth negligible when compared to Earth's effect on me? Looking at the equation above I would think my effect on Earth is the same as Earth's effect on me. Can you help me understand this? Thanks.

You may want to start by reading the current last entry in the FAQ thread in the General Physics forum.

Zz.

 

[granpa]

your own gravitational field is very weak but it pulls on every single atom of the entire earth. that addes up to the same amount of force that the Earth is pulling on you.

you may not see it but when you are pulled toward the Earth the Earth is indeed pulled very slightly toward you. its just a very very tiny amount of motion.

consider 2 atoms. its immediately obvious that one pulls on the other the same as the other pulls on it. yet your body and the body of the Earth are made entirely of atoms. hence the pull is the same.

 

[kbaumen]

Or here's a mathematical proof written neatly in LaTeX:

The Newton's definition of a gravitational force:

$$F_g = G \frac{m_{Earth}m_{you}}{R^2}$$

where G is the gravitational constant ($G=6.6743 \cdot 10^{-11} m^3kg^{-1}s^{-2}$), and by Newton's second law:

$$F = ma$$

therefore:

$$ma = G \frac{m_{earth}m_{you}}{R^2}$$

$$a = G \frac{m_{earth}m_{you}}{m_{a}R^2}$$

If you want to find your acceleration in respect to Earth, you plug in your mass in $m_{a}$, if you want to find Earth's acceleration in respect to your body, you plug in Earth's mass. The force applied to your body and Earth is the same, however, due to these drastic differences in mass, accelerations are different. You accelerate to Earth's surface with ~9.8 $m^1s^{-2}$, while Earth's acceleration is unnoticeable.