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According to Newton's law of universal gravitation, a material point $P$ with mass $m$ attracts a material point $P_0$ with mass $m_0$ with a force directed from $P_0$ towards $P$, of size $k\cfrac{mm_0}{r^2}$ (further for simplicity k=1).

Note that when the point $P_0$ (located at the point $(0, 0, 0)$) of mass $m_0$, is attracted by the point $P_i$ (located in the point $(x_i, y_i, z_i)$) of mass $m_i$, is the force acting between them we express as $$\dfrac{m_om_i}{r_i^2}(x_i,y_i,z_i),$$ where $r_i=\sqrt{(x_i^2+y_i^2+z_i^2)}$.

In the situation where the point $P_0$ is attracted by points $P_1, P_2, \ldots, P_n$ with masses $m_1, m_2,\ldots,m_n$ and the attracting mass has a continuous distribution on the curve $\sigma:[a,b]\rightarrow\mathbb{R}^3, \sigma(t)=(\sigma_1(t),\sigma_2(t),\sigma_3(t))$, the curve is divided into small segments by points $a=t_0<t_1<\ldots<t_n=b$. Then we write the force of attraction as the sum of the vectors

$$\sum_i \dfrac{m_0m_i}{r_i(\sigma(t_i))}(\sigma_1(t_i),\sigma_2(t_i),\sigma_3(t_i)).$$

How to express $m_i$ - the mass of a section from $t_{i-1}$ to $t_i$ using a line integral and the whole as a Rieman sum of a certain integer?

Paull
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  • It is more typically to assume the curve has some parameterization $\gamma: [a,b]\to\mathbb{R}^3$ and a scalar density parameterized along the curve $\rho: [a,b]\to (0,\infty)$ then an infinitesimal force contribution is $\frac{m_o \rho r \mathrm{d}s}{|r|^3}$ – whpowell96 Oct 31 '23 at 14:07
  • I need to understand it in my notation :/ – Paull Oct 31 '23 at 14:24
  • I think all you need is do define a mass density function along the curve so that $m_i = \rho(t_i)~\mathrm{d}\sigma$ and you have your Riemann sum. Be careful with your definition of graviational force in vector form. You should have $F = \frac{m_o m_i}{r_i^3}(x_i,y_i,z_i)$. – whpowell96 Oct 31 '23 at 14:28
  • thanks, I will change $r_i^2$ to $r_i^3$. Can we assume that: $$m_i=\int_{t_{i-1}}^{t_i}f(\sigma_1(t_i),\sigma_2(t_i),\sigma_3(t_i))\sqrt{\sigma_1'(t_i)^2+\sigma_2'(t_i)^2+\sigma_3'(t_i)^2},dt?$$ – Paull Oct 31 '23 at 14:29
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    That looks correct. Approximating that integral into something times $\Delta t$ should give you your Riemann sum for the force. – whpowell96 Oct 31 '23 at 14:33
  • Should I also have $$\sum_i \dfrac{m_0m_i}{r_i(\sigma(t_i))^3}(\sigma_1(t_i),\sigma_2(t_i),\sigma_3(t_i)).$$ instead of $$\sum_i \dfrac{m_0m_i}{r_i(\sigma(t_i))}(\sigma_1(t_i),\sigma_2(t_i),\sigma_3(t_i))?$$ – Paull Oct 31 '23 at 14:36
  • @whpowell96 $$F=\sum_i \dfrac{m_0\cdot \int_{t_{i-1}}^{t_i}f(\sigma_1(t_i),\sigma_2(t_i),\sigma_3(t_i))\sqrt{\sigma_1'(t_i)^2+\sigma_2'(t_i)^2+\sigma_3'(t_i)^2}dt}{r_i(\sigma(t_i))}(\sigma_1(t_i),\sigma_2(t_i),\sigma_3(t_i)).$$ do you know what i need to do more? – Paull Nov 02 '23 at 20:23

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