Question #598b2

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Question #598b2

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Answer 1 · 2017-04-10T10:40:21

At the exact centre of an idealised earth, weight would be zero.

Explanation:

At the exact centre of an idealised (ie perfectly spherical, uniform density, etc) earth, a body would be pulled radially outwards in every direction equally. The net force (weight) would be zero.

To find it I think the quickest approach uses Gauss' Law for gravity:

$\int \int_{S} g \cdot d A = - 4 π G M$ $intint_S mathbf {g} \cdot d mathbf {A} = -4\pi GM$

crucially, where $M$ $M$ is the total mass enclosed within the surface S.

With a spherical Gaussian surface of radius $r$ $r$ concentric with the earth, this means that:

$M = M_{e} {(\frac{r}{r_{e}})}^{3}$ $M = M_e (r/r_e)^3$ where $M_{e}$ $M_e$ and $r_{e}$ $r_e$ are the the earth's mass and radius.

And so the integration becomes:

$g \cdot 4 π r^{2} = - 4 π G M_{e} {(\frac{r}{r_{e}})}^{3}$ $g cdot 4 pi r^2 = -4 pi GM_e (r/r_e)^3$

$\Rightarrow g (r) = - \frac{G M_{e}}{r_{e}^{3}} r$ $implies g(r) = -(GM_e )/r_e^3 r$

$\Rightarrow g (0) = 0$ $implies g(0) = 0$

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Question #598b2

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