4 ms·

While it makes intuitive sense, I find the inverse square law for gravity really hard to comprehend when I think about the solar system, for example Uranus, or

by dahart 7y ago

While it makes intuitive sense, I find the inverse square law for gravity really hard to comprehend when I think about the solar system, for example Uranus, or even the earth, orbiting the sun. When you look at how far apart they are (for example https://news.ycombinator.com/item?id=21735528 https://news.ycombinator.com/item?id=21735528), the radius of the orbits relative to the sizes of the sun & planets, for me it seems to defy the inverse square law that there is still enough force at those distances to keep the planets orbiting. If you built a scale model of the solar system out in the middle of space, would you have the same orbital periods? Like with a scale model of the moon, would a tennis ball sized rock orbit a soccer ball sized rock from a distance of ~25 feet in about a month?

samatman 7y ago
Long story short: yes. To expand a bit: lowering the masses by so many orders of magnitude would make the allowable range of vectors much more precise, and the degree of perturbation which the system could withstand would be much lower. I wouldn't expect such a system to be robust with scale-model comets flying around it, for example.
- dahart 7y ago
  That's interesting; I wouldn't mind hearing a spoiler of the long story. Is there something in the system besides time that doesn't scale relatively? comets at this scale would probably be dust-particle sized, right? Based on the soccer ball / tennis ball scale, my estimate is that a 1km diameter comet would be ~0.02 millimeters in the scale model.
magicalhippo 7y ago
Often overlooked in such representations is just how much larger and more massive the sun is relative to the planets. Uranus is "only" about 20 times further from the sun than the earth, which according to the inverse square law means the suns gravity (and light) is 1/400th of what the earth experiences. But the sun is about 330000 times more massive than the earth, while Uranus is "only" 14 times more massive.
- dahart 7y ago
  Uranus getting 1/400 the gravity that earth does is a good example. And the speed Uranus is moving relative to the sun is not much slower than the speed the earth is moving relative to the sun. Yes, the sun is surprisingly big indeed, so I can appreciate the reminder, but I’m not entirely sure it’s relevant here how big the sun is relative to the planets? The surprise I’m talking about comes from the ratio of the sun’s radius to the radius of the orbit, and that applies to an orbiting mass of any size, from Halley’s Comet to Jupiter. What seems surprising to me is not that the sun is large, it’s that it can attract and keep in orbit something that is at a distance 10,000 times it’s radius (that’s Pluto’s aphelion). The distances between the sun & planets is so much mind-bogglingly larger than the sun that when you see the scale of the sun in context, when you compare the size of the sun to the distance between the sun and the earth, the sun looks incredibly tiny... so tiny that it’s hard to imagine how the gravitational influence is keeping the earth in orbit, especially when thinking about the inverse square law.
tagrun 7y ago
You're looking for Kepler's third law, which is a result of inverse square law and Newton's laws: https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Third_law_of_Kepler https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_m... For all planets, the ratio of the (cubed) distance from the sun to (squared) period is a constant which only depends on the mass of the sun, regardless of the mass of the planets (all assuming that the sun is far massive than any other planet.)

by dahart 7y ago

samatman 7y ago

Long story short: yes. To expand a bit: lowering the masses by so many orders of magnitude would make the allowable range of vectors much more precise, and the degree of perturbation which the system could withstand would be much lower. I wouldn't expect such a system to be robust with scale-model comets flying around it, for example.

dahart 7y ago
That's interesting; I wouldn't mind hearing a spoiler of the long story. Is there something in the system besides time that doesn't scale relatively? comets at this scale would probably be dust-particle sized, right? Based on the soccer ball / tennis ball scale, my estimate is that a 1km diameter comet would be ~0.02 millimeters in the scale model.

magicalhippo 7y ago

Often overlooked in such representations is just how much larger and more massive the sun is relative to the planets. Uranus is "only" about 20 times further from the sun than the earth, which according to the inverse square law means the suns gravity (and light) is 1/400th of what the earth experiences. But the sun is about 330000 times more massive than the earth, while Uranus is "only" 14 times more massive.

dahart 7y ago
Uranus getting 1/400 the gravity that earth does is a good example. And the speed Uranus is moving relative to the sun is not much slower than the speed the earth is moving relative to the sun. Yes, the sun is surprisingly big indeed, so I can appreciate the reminder, but I’m not entirely sure it’s relevant here how big the sun is relative to the planets? The surprise I’m talking about comes from the ratio of the sun’s radius to the radius of the orbit, and that applies to an orbiting mass of any size, from Halley’s Comet to Jupiter. What seems surprising to me is not that the sun is large, it’s that it can attract and keep in orbit something that is at a distance 10,000 times it’s radius (that’s Pluto’s aphelion). The distances between the sun & planets is so much mind-bogglingly larger than the sun that when you see the scale of the sun in context, when you compare the size of the sun to the distance between the sun and the earth, the sun looks incredibly tiny... so tiny that it’s hard to imagine how the gravitational influence is keeping the earth in orbit, especially when thinking about the inverse square law.

tagrun 7y ago

You're looking for Kepler's third law, which is a result of inverse square law and Newton's laws: https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Third_law_of_Kepler https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_m... For all planets, the ratio of the (cubed) distance from the sun to (squared) period is a constant which only depends on the mass of the sun, regardless of the mass of the planets (all assuming that the sun is far massive than any other planet.)