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My teacher would like us to use Newton's law of gravitational attraction to solve a question similar to this one:

"Two objects exert a force of gravity of 20N. One object's mass is increased by a factor of 5 and the masses are moved 6 times closer together. What force strength do they exert on each other now?"

This question is different from the one that my teacher assigned me so I can do that myself. My teacher would like us to solve it using an equation. Is he referencing to the universal gravitational constant to solve the question? How would you solve it?

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closed as off-topic by ACuriousMind, user10851, AccidentalFourierTransform, John Rennie, David Z Apr 4 '16 at 7:53

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Use that the gravitational force between two (point) masses $m_1, m_2$ separated by a distance $r$ depends on the product of the masses $m_1 \times m_2$ and is inversely proportional to the square of the separation, $\frac{1}{r^2}$. What you have to do is figure what happens if one of the masses goes from $m_1 \to 5 m_1$ and $r \to \frac{r}{6}$, so you don't really need to use the value of $G$.

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The formula you are looking for is the one for gravitational force $$F_g = \frac{Gm_1m_2}{r^2}$$

since $G$ and $m_2$ do not change in your example, we can write this more concisely as $$F_g \propto \frac{m_1}{r^2}$$ You will then find the answer by replacing $m_1$ with $5m_1$ and $r$ with $\frac{r}{6}$.

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  • $\begingroup$ oh I see. So you just omit the G since it's not used. $\endgroup$ – IronZeppelin Apr 3 '16 at 23:06
  • $\begingroup$ Note that the second relationship is a proportionality and not an equality which is why you can ignore constant terms. $\endgroup$ – hsnee Apr 4 '16 at 0:10
  • $\begingroup$ The answer would be better if you simply referenced the proportionality relationship and let the OP actually solve the problem. Try to avoid providing a complete solution to homework-style problems. $\endgroup$ – Bill N Apr 4 '16 at 1:36
  • $\begingroup$ @IronZeppelin - No, the G is used in both equations. It's just that it's the same in both equations, so you can ignore when comparing the two. When you take the ratio of two quantities, common factors drop out - that is, the ratio of G/G is 1. $\endgroup$ – WhatRoughBeast Apr 4 '16 at 2:02

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