How do Earth's gravity and a pencil's force compare according to Newton's laws?

[krackers]

How would the size of the force exerted by Earth’s gravity on an object (lets say... a pencil) compare with the size of the force the pencil exerts on the Earth? I'm pretty sure that the forces would be equal from Newton's third law but it could also be that the Earth, with a larger mass, exerts a larger force on the Earth compared to the pencil, which has a smaller mass and thus exerts less force.

Searching for answers [http://wiki.answers.com/Q/What_is_larger_the_force_the_earth_exerts_on_you_or_the_force_you_exert_on_the_earth] seem to give two varied results: either the body with a larger mass (earth) will exert a greater force on the smaller mass (pencil) than the smaller mass (pencil) exerts back on the larger mass (earth), or that the forces will be equal due to Newton's Law of Gravitation and Newton's 3rd Law.

 

[Doc Al]

How would the size of the force exerted by Earth’s gravity on an object (lets say... a pencil) compare with the size of the force the pencil exerts on the Earth? I'm pretty sure that the forces would be equal from Newton's second law but it could also be that the Earth, with a larger mass, exerts a larger force on the Earth compared to the pencil, which has a smaller mass and thus exerts less force.

The gravitational force they exert on each other is equal and opposite.

 

[Drakkith]

The gravitational force they exert on each other is equal and opposite.

Can you elaborate? I thought the Earth exerted a far greater force on the pencil than vice versa.

 

[krackers]

The gravitational force they exert on each other is equal and opposite.

But isn't that the gravitation force between the two objects? That is the force represented by F=G * (m*M)/r^2

In that case, you would be right as when the Earth pulls on the pencil then the pencil would equally pull on the Earth by Newton's 3rd Law. In other words, the action and reaction force would be equal.

However, what happens when you just have one object pulling on the other? The force at which the Earth pulls the pencil in comparison to the force at which the pencil pulls the earth. The pencil would have a smaller gravitational field due to its smaller mass and the Earth would have a larger gravitational field right? So in that case wouldn't pencil would pull on the Earth with less force than the Earth pulls on the pencil?

 

[1977ub]

Can you elaborate? I thought the Earth exerted a far greater force on the pencil than vice versa.

The Earth causes a far greater acceleration on the pencil than vice versa. a = F / m.

 

[krackers]

The Earth causes a far greater acceleration on the pencil than vice versa. a = F / m.

But the force exerted by the pencil on the Earth and the Earth on the pencil is the same?

 

[1977ub]

http://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Gravitational_Forces

Thus gravitational force exerted by a sphere of mass M on another sphere of mass m is given by the following formula: F_{grav} = \frac{GMm}{r^2}

it's the same force whether you think of M as larger or smaller than m.

 

[TurtleMeister]

But the force exerted by the pencil on the Earth and the Earth on the pencil is the same?

That is correct. The gravitational forces are equal and opposite. To understand it better, try thinking in terms of momentum. If the Earth and the pencil are in free fall toward each other (no atmosphere), the momentum of the Earth will be the same as the momentum of the pencil at any given instant. |M_e * A_e| = F = |M_p * A_p|.

 

[Drakkith]

Is this because while the pencil exerts very little force, the Earth has a LOT more mass for that small force to act on, while the opposite happens for the Earth on the pencil?

 

[Chestermiller]

Is this because while the pencil exerts very little force, the Earth has a LOT more mass for that small force to act on, while the opposite happens for the Earth on the pencil?

If you change the word "force" to "gravitational field" in this statement, it is right on target.

 

[WannabeNewton]

Is this because while the pencil exerts very little force, the Earth has a LOT more mass for that small force to act on, while the opposite happens for the Earth on the pencil?

The pencil exerts the exact same gravitational force on the Earth as the Earth exerts on the pencil, they lie along the same line, and are opposite in direction. The difference is that the pencil accelerates a lot more than the Earth does.

 

[Doc Al]

But isn't that the gravitation force between the two objects? That is the force represented by F=G * (m*M)/r^2

Yes. The force is an interaction between two objects.

In that case, you would be right as when the Earth pulls on the pencil then the pencil would equally pull on the Earth by Newton's 3rd Law. In other words, the action and reaction force would be equal.

Yes.

However, what happens when you just have one object pulling on the other?

Can't happen. If A pulls B, then B pulls A.

The force at which the Earth pulls the pencil in comparison to the force at which the pencil pulls the earth. The pencil would have a smaller gravitational field due to its smaller mass and the Earth would have a larger gravitational field right?

Yes, if you consider the field due to each object, the pencil's field is much weaker. But the force they exert on each other is the same.

So in that case wouldn't pencil would pull on the Earth with less force than the Earth pulls on the pencil?

No.

 

[1977ub]

Yes, if you consider the field due to each object, the pencil's field is much weaker

Is there an equation for that?

 

[jtbell]

$$F = G \frac {mM}{r^2}$$

so if we take m as the mass of the pencil and M as the mass of the earth, then the gravitational force of the pencil on the Earth can be written as

$$F = M \left( G \frac{m}{r^2} \right)$$

where the part in parentheses is the gravitational field of the pencil. Or going the other way, the gravitational force of the Earth on the pencil can be written as

$$F = m \left( G \frac{M}{r^2} \right)$$

where the part in parentheses is the gravitational field of the earth. Recalling another formula for gravitational force, F = mg where g is the "gravitational acceleration" at the Earth's surface (9.8 m/s²), we can see that we can also think of g as the gravitational field at the Earth's surface,

$$g = G \frac{M}{r^2}$$

with units N/kg (Newtons/kilogram), which is equivalent to m/s². In this case r is the radius of the earth, i.e. distance from its center.

 

[1977ub]

we can see that we can also think of g as the gravitational field at the Earth's surface

oh yes of course. so pencil's field is G * m / r^2

if pencil were spherical.

 

[ZapperZ]

Note that there is a related entry in the FAQ on this:

https://www.physicsforums.com/showthread.php?t=511172

WannabeNewton has correctly explained that while the force acting on each object is of the same magnitude, the acceleration experienced by each of them are not. One can show this easily using a similar step shown in the FAQ.

Zz.