Expert: Philip A. Stahl - 1/19/2011

Question
QUESTION: Hi Doctor Stahl,
This is a bit of a thought experiment question, hope it makes sense,  If a void could be created at the exact centre of a planet or star, would the surrounding upward forces cancel out the weight any observer, making them experience 0 G?

ANSWER: Hello,

I'd have to say that a 0 G experience would not be so produced, given that no matter how large your 'void' you will still have mass for the planet, and this is what determines weight.

Thus,

mg (weight or the force of gravitational attraction)

= G M(P) m/ r^2

where G is the gravitational constant and m the reference mass - say of a person, with r the distance to the center of the planet's mass, M(P).

Thus, g = G M(P)/ r^2

so, unless your void is such that M(P) = 0, this isn't going to occur!

But, if M(P) = 0 you no longer putatively have a planet!

---------- FOLLOW-UP ----------

QUESTION: Yes I agree it would no longer be a planet I suppose the object that I would have would be something closer to a Dyson Sphere (without the star at the centre)...

Apologies for this should of perhaps stated some assumptions:

1. the observer is at the exact centre of this void.
2. The spherical shape is uniformly dense and smooth both inside and out

I think the gravitational geometry might be different from the spherical equation your stating for 2 reasons:

-in the case of the model

g = GM(P)/R^2

the radius you are using would have to have the value of 0 the exact centre this would cause the value of g to be infinite... since r^2

this is clearly not case since we are not dealing with a typical sphere and the centre of mass would be elsewhere from the centre of the "sphere shape"

And

if g does does have a value which direction would I be attracted to?

I think since r (or something similar to r) would be the distance of rock (or whatever matter the shape is constructed of) between the observer and the outer surface of the object (i.e. The thickness of the shell), plus the distance from that to the centre of the object (but in every direction)

do you know where I could find out if Dyson or some else did an equation for working out the forces inside his spheres?

Answer
Hello,

Okay- thanks for clarifying your question. Several problems here: first, the definition of "weight" (mg) as equal to the force F of gravitational attraction, is predicated on the existence of a reaction force. Thus, for a book lying atop a table, the weight of the book is only measurable because TWO forces act in concert: a downward force W= mg and a normal force (N) upwards. If there is no possible N, there can be no W. This is why a scale works, because an upthrust or normal force acts against your downward force (W).

In your scheme, if the observer is located at the center of a void, then clearly he has no support (N=0), hence W = 0. (It doesn't matter if r = 0, for the reason given above, so it isn't a case of "infinite g"). In effect, being inside a "void" would mean the observer or person cannot be in any equilibrium, hence will fall freely - just as if he were in an elevator (lift) in which the cables were cut.

In the case of free fall, then g = 0. Again, the person is weightless since he has no support for the same reason a person trapped in a freely falling lift has no support. Hence, weightless.

Sorry, I can't be of help Re: Freeman Dyson, but in any case what you're describing isn't really a "Dyson sphere" but something else entirely.

See, e.g.:

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


Footnote: Actually, for the uniform shell sort of hollow sphere you describe, the center of mass would be at its very center, in much the same way that the center of mass of a doughnut is located at its geometric center - not anywhere in its substance. A center of mass need not actually be in a mass to be the c.m., in other words.

Hope this helps!