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重力增加接近核心你得到了什么?

还是上面的地幔和地壳中你抵消提高一度,它实际上减少?

回答

这是一个同伴(其他答案)(//www.hoelymoley.com/a/19135/18801)。其他答案给实际上加速度值* *是基于知识的组成的星球。在这个答案我想介绍一些数学工具允许你原因重力变化如何和周围的行星。(David Hammen的答案)(//www.hoelymoley.com/a/19165/18801),我认为他写的第二部分的重叠,也有版本中的一些数学的第二部分,是非常值得一读。# #一个理想化的(错误的)地球地球首先考虑一个理想:地球在这个模型中假设地球是球形;——而不是旋转;——恒密度。这些都是相当不错的前两个近似。第三是* *不是一个好的近似:地球的密度随深度变化显著。这个简化的模型我们可以使用一个著名的结果,证明了牛顿,称为壳定理(https://en.wikipedia.org/wiki/Shell_theorem)。 This says two things: 1. a spherically-symmetric mass distribution can be treated gravitationally outside it as if it was the same mass concentrated at a point at the centre of the sphere; 2. a spherically-symmetric *shell* of mass (so a hollow sphere) has *no* gravitational influence on any body *inside* the shell. From this it immediately follows that, at any point *inside* our model Earth, you can ignore all the mass *outside* that point and then just consider the gravitational influence of the mass *inside* the point, and we can also consider that mass to be concentrated at the centre of the model Earth. So, well, let the density of the planet be $\rho$. Then the mass within a radius $r$ is $m(r) = 4\pi r^3 \rho/3$, for $r\le R_E$, where $R_E$ is the radius of the Earth. And using Newton's law of gravitation, we get the acceleration due to gravity, $g(r)$ to be $$\begin{aligned} g(r) &= \frac{Gm(r)}{r^2}\\ &=\frac{4\pi G\rho r}{3}\quad r \le R_E \end{aligned}$$ Which shows you that $g$ reduces linearly as you approach the centre of the Earth in this model. ## A more realistic model Earth A nice thing is that you can use most of the above even if $\rho$ is not constant: so long as $\rho = \rho(r)$, only, then you can still use the shell theorem to compute $g(r)$, but you now have to do some integral over $r$ to compute $m(r)$. So you can relax the third requirement above, at the cost of a little more work. I believe that, in practice, $\rho$ does depend only on $r$ to a fairly good approximation, so this model will let us compute how gravity actually does vary. In the case where $\rho = \rho(r)$ we can use the shell theorem to write $g(R)$ at some radius $R$ (I'm using $R$ not because I want to integrate over $r$). $$\begin{aligned} g(R) &= \frac{G m(R)}{R^2}\\ &= \frac{G}{R^2}\int_0^R 4\pi r^2 \rho(r)\,dr \end{aligned}$$ Where the integral is the mass of the planet to radius $R$. It's clear from this that $g$ is constant if $\rho(r) = \rho_0/r$: a planet with a density which goes down as $1/r$ has gravitational acceleration which is constant throughout its volume. Note that this planet has infinite density at its centre: since that's never true for planets – we're not dealing with black holes here, and if we were we could not use Newtonian gravitation – this means that for all real planets, $g \to 0$ as $r \to 0$. More generally we can differentiate the above to get an expression for $dg/dR$: $$\frac{dg}{dR} = 4\pi G\rho(R) - \frac{2G}{R^2}\int_0^R 4\pi R^2\rho(r)\,dr $$ Defining $\bar{\rho}(R)$ as the mean density of the planet to radius $R$, then the integral in the second term is just $4\pi R^3\bar{\rho}(R)/3$, so we get $$\frac{dg}{dR} = 4\pi G\left(\rho(R) - \frac{2}{3}\bar{\rho}(R)\right)$$ This tells you how $g$ behaves with depth, and in particular whether it goes up or down with depth for a general spherically-symmetric mass distribution, if you know the radius, the density at that radius, and the mean density inside that radius (or equivalently the radius, the density, and the mass inside that radius). So, if we can measure how the density of Earth varies with depth, and that it varies *only* with depth to a good approximation, then this model can be used to compute how $g$ varies with depth, to a good approximation.

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\ begingroup美元这头牛有点太球形。最好是显示在何种情况下重力加速度增加或减少与中心距离增加,假设球质量分布。一个统一的质量分布没有意义的一个行星大小的物体(即使是一个很小的星球)。 \ endgroup美元
- - - - - -大卫Hammen
2020年2月7日在4:39
1

\ begingroup美元 @David Hammen我不懂你的评论。一方面,你说“最好....假设球质量分布”。然后你说“一个统一的质量分布没有意义……”。的区别是什么? \ endgroup美元
- - - - - -乔治Menoutis
2020年2月7日,8分到达
1

\ begingroup美元 @GeorgeMenoutis第一个意味着密度取决于半径。由于行星的密度均匀,线性下降不是一个有用的近似,但球对称星球的数学不是更加困难,它将更接近真相。 \ endgroup美元
- - - - - -JiK
2020年2月7日12:56
1

\ begingroup美元尽管任何反对或谬论,这个问题,这些答案都是伟大的,所有的加在一起,因为他们显示不同的模型(质点在中心,统一的密度,密度变化)建模得到不同,越来越多的准确的结果。 \ endgroup美元
- - - - - -大卫·康拉德
2020年2月7日15:48
1

\ begingroup美元 @SemidiurnalSimon:在当前版本是更明确的初始理想化是错误的,第二部分处理一般球对称的质量分布。最后它还明确表示,它的存在提供参考你需要的数学工具,它应该总是说。 \ endgroup美元
- - - - - -user18801
2020年2月12日11时20分

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