When I studied physics, one of my professors always had us do dimensional analysis to see if we could come up with a rough guess doing problems like this.
The time that it takes for someone to fall into the sun is related:
Inversely to the distance from the sun (D)
Inversely to the gravitational constant (G)
Directly to the mass of the sun (M)
That’s it. It doesn’t depend on the your mass (we can thank Galileo for that tidbit), or anything else.
So we need to combine those items in such a way that we get “time”.
G has units: d^3/t^2/m
M has units: m
D has units: d
So, it’s got to be sqrt(D^3/GM).
Fortunately, we know that M = (4pi^21AU^3)/(G1yr^2) (note: 1AU = D; how lucky!)
Even simpler, John Mace. Given all of those constants, we already know of at least one timescale we can derive: The orbital period of an object given those parameters, and the simplest orbit, is one year. So right off the bat we know that the answer is going to be one year, times some mathematical constants (which are usually not too far from 1; within an order of magnitude or so).
We can break it down even further: An object in a circular orbit takes 1/4 of its orbital period, not the whole period, to go from X = 1 AU to X = 0. So our next estimate would be 1/4 year. And we can refine that yet further: In a circular orbit, the magnitude of the force never changes (and in fact the X component of the force decreases), but in the straight-in fall, the magnitude of the force increases. So this tells us that the quarter-orbit estimate is too long, and that therefore the correct answer should be somewhat less than 1/4 year.
So if I’m following you correctly, Sedna has a “year” that is 10500 earth years, so something in free fall from it’s furthest point of orbit would be:
10500/2*3.14 = 1671 years.
Did I do that right? Or does the elliptical orbit of Sedna screw it up?
Now, iamnotbatman, you came up with a velocity of 1.4 million miles per hour at the point of contact with the sun, assuming a free fall from earth radius. Am I reading you correctly that it doesn’t really matter if one falls from Earth or Sedna, you’re still going to splat into the sun at more or less the same speed?
I would guess you did that right. Basically, the reason it takes so long is the same reason you get most of the bang for your buck falling from earth: the force is far stronger inside the earth’s orbit than outside. So you start really, really slowly. And, of course, it would take an infinite time from infinitely far away.
Follow-up question: when we say that the gravitational acceleration of earth is 9.8m/s^2 I take it we mean 9.8m/s^2 at points close to the earth’s surface. So are we saying that something, say, at the moon’s orbital radius would actually accelerate more slowly than that? So in other words the acceleration itself accelerates? Is there a formula that captures the rate of change of a rate of change?
ETA: I didn’t see SMT’s post befor I wrote this and the airplane doors just shut so I have to put my iPhone away!
As far as a formula goes, you are talking about what is generally talked about as a tidal force: the change in the force with distance. The tidal force will decrease as one over the cube of the distance, as opposed to one over the square.
That is a big difference. The moon and the sun, for example, both pull on the earth. A quick google, which is faster than a calculator, shows that the gravitational force of the sun on the earth is 180 times that of the moon, but they contribute pretty equally to the tides.
It’s probably misleading to phrase it that way. Acceleration means that speed changes with respect to time. What we’re looking at here, though, is an acceleration that changes with position. And yes, of course there are formulas for this, dating back to Newton himself.
I worked the numbers out a while back in this post. But yes, the sun pulls on the Earth almost 200 times what the moon does, but has a bit under half the tidal effect.
Yeah - when I calculated the period with which a Niven Ringworld would oscillate if you bobbed it up and down was 377 days - only slightly longer than a year because the Ring is a bit wider than 1 au, and the Ring is massive enough that its mass can’t be neglected.
If you picture the motion to be a circle on the X-Y axis with radius 1 AU, then when the earth is at X=1, you can see that Y=0. Go a quarter of the way around, then X=0, and Y=1. The point is to think of the earth as “falling” from X=1 to X=0, ignoring the Y-coordinate.
So here’s something I’ve been wondering for a while. So you have two bodies, A and B. If you drop body A onto body B, you use B’s mass and the distance between the two bodies to figure out the force of gravity and how quickly A accelerates towards B, ignoring A’s mass.
If you drop body B onto body A, you use A’s mass and the distance between the two bodies and do exactly the same thing, ignoring B’s mass.
But in actuality, both of these are happening at the same time in any “drop it and see how fast it goes” scenario. A is pulled toward B, and B is pulled toward A. The effect is immeasurably small with something like dropping a bowling ball on the earth, but if you consider a universe that only consists of two identical earths, they will crash into each other at the point halfway between them - the standard “ignore the mass of the falling body” treatment will have the two earths collide either all the way over at one earth or the other, depending on which one you treat as the falling body.
Now having said all this, what happens if you drop the earth into the sun? How fast will the earth be moving towards the sun, and how fast will the sun be moving towards the earth?
What type of fun gravity effects can we expect? For example is there a point where the sun’s gravity overwhelm’s earth’s and everything on the planet “falls up” towards the sun? Will earth eventually be accelerating faster than the 9.8m/s^2 that earth’s gravity produces, so people on the night side of the planet “fall up” as earth is sucked away from them?
No, due to the equivalence principle, the fact that the earth will be in free-fall toward the sun means that no one on the planet will feel any of the effects of the sun’s powerful gravitational field (tidal effects will be pretty small, I think). So there won’t be any fun gravity effects: yes, you will be accelerating towards the sun, but not any faster than the floor beneath your feet!
About your other question: the sun will not be moving much towards the earth, because the sun is over 300000X more massive than the earth. The earth is sort of like the bowling ball you mentioned.
Actually, you always have to take both masses into account, unless one is so much more massive than the other that it doesn’t matter to the accuracy of solving the problem. (A bowling ball falling towards the earth.)
It might help to remember that an orbit is a free fall. The earth just keeps missing the sun, because of its velocity orthogonal to the direction towards the sun (and in the plane of the orbit). Of course, the sun is much, much bigger than the earth. For all practical purposes, the earth revolves about the sun. (Actually, GPS is a practical purpose for which the earth’s mass relative to the sun is important, IIRC.)
The earth and moon revolve about their center of mass. The center of mass is within the earth, but most of the way towards its surface. Both objects are perpetually falling towards that center of mass but, Adamseque, they just keep missing.
The vis-viva equation is v^2 = G(M1+M2) (2/r-2/a) where v is the relative velocity between two objects (one of mass M1 and the other of mass M2), r is the distance between them, and a is the semimajor axis of the elliptical orbit that either object is on, if you consider the other object to be fixed - but when one of the objects is the Sun, and one is a planet, or when one object is a planet and the other is a satellite, you can neglect the smaller mass entirely and get a good enough answer. For your two identical Earth case, though, you can’t ignore either mass - so you get a velocity that is 41% faster than it would be for the Earth/bowling ball case.
