Gravitational potential energy
𝑟 2 , so does lobe #2, and so does the system of the two lobes, because they all represent the same. Thus, the energy liberated by collision is 𝐺𝑚 1 𝑚 2 /𝑟 1 +𝑟 2 .
We can visualize the work done by gravity by sketching a force against distance graph, and record the process when they are pulled together from infinitely far distance. To simplify the matter, we may let the two lobes be of equal mass. The distance, 's', can be the coordinate of both mass #1 and #2, measured from the initial position of mass #1. The 's' line will look like this:
Because the two masses are identical, the line should look symmetrical. 'x' is an infinite distance. Consider two cases:
case 1: the two masses freely attract each other until they collide (distance 'R' apart). case 2: mass #1 is stationary, mass #2 is attracted until it collides (distanced 'R' from #1).
For case 1, the graph should look like the one on the left. It shows the force they feel before collision in relation to their 's' coordinates. The curve above 's' axis is mass #1, and the curve below 's' axis is mass #2, because we define the force as positive in the right direction. The total area between the curves and 's' axis (call it A1) is the work done by gravity.
𝑐𝑎𝑠𝑒 1
𝑐𝑎𝑠𝑒 2 (𝒐𝒏𝒍𝒚 𝑡ℎ𝑒 𝑛𝑒𝑤 𝑎𝑑𝑑𝑒𝑑 𝑐𝑢𝑟𝑣𝑒)
The only curve of case 2 is the curve newly drawn. I have put it within the previous graph because it is easier to compare. In case 2, mass #1 has no curve because it is not moving. The total work done by gravity is the area between the new curve and 's' axis (A2). The new curve spans between 𝑠 = 𝑅 and 𝑠 =2𝑥 +𝑅 , therefore twice the displacement than the curve of #2 in case 1 (call it the previous curve). Given that the new curve is just 'the previous curve' stretched by a factor of 2 on this 's' axis, the area enclosed should also be 2 times larger. It is easier to see by rotating the graph by 90 degrees anticlockwise. However, if one wishes to find the mathematical derivations of the exact areas enclosed by the graphs, here are my procedures:
𝐺𝑚 2 (2𝑠 − 2𝑥 − 𝑅) 2
𝐹 𝑝 =
𝐹 𝑛 = 𝐺𝑚 2 𝑠 2 𝐴 𝑝 =∫ 𝐹 𝑝 𝑑𝑠 2𝑥+𝑅 𝑥+𝑅 𝐴 𝑛 =∫ 𝐹 𝑛 𝑑𝑠 2𝑥+𝑅 𝑅 We will find that 𝐴 𝑛 =2𝐴 𝑝
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