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My professor said that since mass is directly proportional to the volume, if we triple the radius we would consequently increase the mass by 27 times. Afterwards we would just plug the numbers newtons law of gravitation formula. The final part of the problem requires the gravitational force formula:

\begin{aligned} F&=\ \frac{GM_{1}M_{2}}{R^2} \\ \\ F_{R} &=\ \frac{G(M)(M)}{R^2} \\ \\ F_{3R} &=\ \frac{G(27M)(27M)}{(3R)^2} \\ \\ &=\ \frac{G(27M)(27M)}{9R^2} \\ \\ &=\ \frac{81G(M)(M)}{R^2} \\ \\ &=\ 81F_{R} \end{aligned}

​​ ​​ However I don't understand why he placed R instead of 2R for the second step. Shouldn't it be 2 R because R should be the distance from the center of mass? Same thing with the third step. Shouldn't R be 6R? enter image description here

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Shouldn't it be 2 R because R should be the distance from the center of mass? Same thing with the third step. Shouldn't R be 6R?

Yes, it should. This is a case where two wrongs made a right. Your instructor made the same mistake twice, and those mistakes happened to cancel. Use $\frac {GM^2}{(2R)^2}$ and $\frac{G (27M)^2}{(6R)^2}$ and you'll get the same ratio, an 81-fold increase in the gravitational force.

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The force of gravity varies as the inverse square with distance from the object. The leftmost point of the right sphere will feel a stronger gravitational attraction to the left sphere, than would the right sphere's central point or leftmost point.

The variation in gravitational force across a body is known as a tidal force (named for the tides; an affect of the moon's tidal force on the earth).

You will get the same answer regardless of what point you consider. Try the rightmost, centre, and leftmost. Each are valid.

Edit: The accepted answer is incorrect. There is no mistake in the original solution. It doesn't matter where you choose the reference point to be.

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