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Interestingly, I had this question after watching the first Austin Powers movie. I tried searching this question on other sites, but I couldn't find a conclusive, mathematical answer.

Dr. Evil threatens to detonate a nuclear weapon at the center of the Earth, causing much destruction to the planet. I was curious about the impact this might have. I first assumed that an explosion's force would follow the inverse square law, so I could isolate the force applied on the barrier between the outer and inner core of the Earth.

While rudimentary, I figured that I could calculate the gravitational force at the center, and then use a standard

$$\sum \text{Force} = F_{\text{explosion}} - F_{\text{gravity}}$$

However I ran into the problem of applying Newton's universal law of gravitation at the center of a body. No matter how you evaluate this, the center of mass of any ring or spherical body is at the same spatial location as a "point" at the exact middle, where the explosion would occur.

$$F = G\frac{m_1 \, m_2}{r^2}$$

We can safely assume that at the said point, $F_{\text{net}} = 0$, as something there wouldn't move.

Is this strictly a question of pressure? But if so, that pressure would be caused by gravity itself (the extreme heat at the Earth's core nonwithstanding).

I'd appreciate any insight you guys have into this problem, as well as any critiques of my assumptions or initial methodology.

P.S - if two bodies are infintesimally close together, wouldn't the force between them be infinite? (Sorry too lazy to make a limit in LaTeX here)

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    $\begingroup$ Dr Evil is evil but not very smart. The centre of the Earth would be the least damaging place for a nuclear explosion to occur. $\endgroup$
    – my2cts
    Commented Aug 13, 2021 at 8:13
  • $\begingroup$ In non relativistic regime, $F_{net}=0$ at the centre . In extreme pressure condition, relativity suggest that pressure would also contribute to gravitational field. So Newton's equations will be inappropriate, instead you will have to solve the Einstein's field equations $\endgroup$
    – paul230_x
    Commented Aug 13, 2021 at 8:40
  • $\begingroup$ The "universal" law of gravity only applies outside bodies, and far away also. $\endgroup$ Commented Aug 13, 2021 at 12:01

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The gravitational force at the centre of a spherical mass vanishes.

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The acceleration due to gravity at a distance $r$ from the middle of the earth, depends only on the matter within the sphere of radius $r$ - so assuming constant density $\rho$ of the earths core it's

$$G {\frac{(4/3)\pi r^3\rho }{r^2}} = \frac{4\pi G r\rho}{3} $$

and infinities are avoided, the acceleration approaches zero as $r$ approaches zero.

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Answer for the question in "P.S." part:

Newton's Law at $r \rightarrow 0$ is not determined. It is believed that new type of theory (Quantum Gravity) has to be applied at these tiny scales. Moreover, don't forget about Heisenberg's principle, which prohibits us from diving into that tiny distances.

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    $\begingroup$ QG has (for practical purposes) nothing to do with calculating gravitational field at centre of Earth ($r\to 0$). Classical theory doesn't blow up at centre of Earth. $\endgroup$
    – paul230_x
    Commented Aug 13, 2021 at 8:30
  • $\begingroup$ sorry, didn't mention. My answer is related to the last question $\endgroup$ Commented Aug 13, 2021 at 8:40

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