Gravitional Force On Composite Body

[Bashyboy]

Why is finding the gravitational force on the center of mass of a composite body the same as finding the gravitational force on each individual constituent of the composite body?

 

[geologic]

If the body is modeled such that the acceleration felt by each particle is the same, then

Ʃm_ia_i= M Ʃ m_i / M a_i = M d²/ dt² Ʃ m_i/M x_i = M a_cm = M g

M is the total mass and Ʃ m_i/M x_i is by definition the center of mass. Assuming they all feel the same field, the result follows.

A case where the field is non-uniform can provide different results. Imagine a field finite at the origin and zero elsewhere. If two masses are at +/- 1, they feel no field, but there is a finite field at the location of their center of mass.

 

[WannabeNewton]

This can be stated more generally. The force on the composite body is given by $F = \sum m_{i}\ddot{r}_{i} = \frac{\mathrm{d} ^{2}}{\mathrm{d} t^{2}}\sum m_{i}r_{i} = M\ddot{R}$ where $M$ is the combined mass of the particles and $R$ is the position vector of the center of mass.

EDIT: Oops, I was typing this up whilst geologic replied so sorry for the repeat!

 

[Bashyboy]

Geologic, why are you proposing such a specific example? I am having a difficult time seeing how the situation you provided is applicable.

 

[D H]

Why is finding the gravitational force on the center of mass of a composite body the same as finding the gravitational force on each individual constituent of the composite body?

It is not the same. Consider a very long, very slender vertical rod such as the hypothetical space elevator. The gravitational force on this rod is *not* the same as the force on an equivalent point mass located at the center of mass of the rod.

In general, center of gravity ≠center of mass. In practice, the distinction is only meaningful for very long, very slender objects. In most cases, the difference between the two is ridiculously small.

 

[Bashyboy]

So, problem 10 in the link http://physics.usask.ca/~pywell/p121/assignments/Solutions-11.pdf , assumes that the center of mass is the center of gravity?

 

[D H]

Strictly speaking, yes, what they did is invalid. Practically speaking, what they did is okay. The difference between center of mass and center of gravity in that problem is a whopping 1.25 millimeters.

Another way to look at it: The calculated force is expressed using three digits of precision. Their result using center of mass = center of gravity is just fine to three or even four places of accuracy. The distinction doesn't rear its ugly head in this case until you get to the fifth decimal place.

 

[dipole]

It is not the same. Consider a very long, very slender vertical rod such as the hypothetical space elevator. The gravitational force on this rod is *not* the same as the force on an equivalent point mass located at the center of mass of the rod.

In general, center of gravity ≠center of mass. In practice, the distinction is only meaningful for very long, very slender objects. In most cases, the difference between the two is ridiculously small.

In more general terms, the center of gravity coincides with the center of mass only for uniform gravitational fields. If the field varies in space, then in general the center of mass and of gravity will be different.