# How long would it take the Earth to fall into the Sun?

*If the Earth suddenly stopped orbiting the Sun, I know eventually it would be
pulled in by the Sun's gravity and hit it. How long would it take the Earth to
hit the Sun? I imagine it would go slowly at first and then pick up speed. *

It would take about two months for the Earth to hit the Sun (and yes, you are right; it would go slowly at first and pick up speed as it continued to fall).

How did I get this number? There are three ways to get it: one is to calculate it exactly (which is long and boring and involves integral calculus), another is to use a neat trick involving orbital physics (explained further below), and the third is to do what is known as a "back of the envelope" or "order of magnitude" estimate, where you use your intuition to make some simplifications to the problem that allow you to approximately guess the answer without using any messy mathematics.

Back of the envelope estimates are used all the time by physicists and astronomers, and in this case (like many others), the back of the envelope estimate does an extraordinarily good job! Therefore, I think it's worth going over in some detail.

The simplest way to start a back of the envelope estimate is to make a list of all the things that might influence the answer to the question. As a silly example of what I mean, it's easy to think of some things that would *not* influence the answer: the winning lottery number in New Jersey six months ago, the weather in Switzerland the day the Earth starts falling, etc! But there are several things that conceivably could influence the answer, and among these might be some properties of the Sun and Earth as well as of the force of gravity which pulls the Earth in. So we need to make a list of these things. (If you want to try this yourself, go ahead and make a list now before reading further.)

There are two obvious things that should be on the list:

- The mass of the Sun (which we'll call
**M**); if the Sun were hypothetically made more massive, its gravitational pull on the Earth would be stronger, and the Earth would take a shorter amount of time to fall. The mass of the Sun is known to be around 1.99 x 10^{33}grams. - The distance between the Earth and the Sun (which we'll call
**R**); if this distance were hypothetically increased, then the Earth would have farther to travel before it hit the Sun, so it would take a longer amount of time to fall in. The distance between the Earth and Sun is known to be around 1.50 x 10^{13}centimeters.

There's a third thing that influences the answer to the problem, although this one might be less obvious if you don't have a physics background:

- Newton's gravitational constant (which we'll call
**G**); this is a physical property of the universe which determines how strong gravity is compared to other forces. Its value is known experimentally to be 6.67 x 10^{-8}(in the rather funny units of "centimeters cubed per second squared per gram"). It's a safe bet that in any physics problem which involves gravity,**G**is going to pop up somewhere, so we'll include it here.

Can you think of anything else that might influence the answer to the problem, besides the above three quantities (**M**, **R** and **G**)? I can't! At the bottom of this essay, I'll mention a couple other possibilities and explain why they aren't important, but for now, suffice it to say that we can roughly calculate the time it takes the Earth to hit the Sun using only those three numbers.

The fascinating thing about back of the envelope calculations is that once you've determined the important quantities, you're pretty much finished thinking about the problem! Because we know that our final answer must have units of time (e.g., seconds, minutes, months, etc.), all we really need to do is find a way to mathematically combine **M** (which has units of mass), **R** (which has units of distance) and **G** (which has units of "distance cubed per time squared per mass") in such a way as to give units of time, because the final formula we are seeking will *have* to have these three quantities combine in such a way to result in an answer that has units of time.

If you'd like, you can play around with the three quantities and try to find a formula that gives units of time. It might help to remember the physical intuition we came up with above, where we argued that increasing **G** or **M** (i.e., increasing the gravity) would shorten the time it takes the Earth to fall, while increasing **R** would lengthen the time. In order for this to work, **G** and **M** (raised to some power) must appear in the denominator of our final equation, while **R** (raised to some power) must appear in the numerator.

If you play around with the three quantities, you'll eventually find that the only formula which satisfies the above constraints is:

Time = ( R^{3} / GM)^{1/2}.

(Here, the superscript "1/2" means that the square root of the term in parentheses is being taken.) This is the formula we will use for our back of the envelope estimate. Is it necessarily the correct formula? No, not exactly. There might be a number (like 2, or 3, or 7) that multiplies this formula out in front, and there is no way we can determine what this number is without doing a complete calculation. However, in simple problems like this one, the number out in front is not likely to be very large; astronomers like to say that the above equation is valid to an "order of magnitude." It may not give us the *exact* right answer, but it will give us one that is close enough for estimation purposes.

If we plug in the values of **M**, **R** and **G** to this formula, we get an answer of 5 million seconds, which is about 58 days. How close is this to the real answer? Well, if you go through all the complicated integral calculus needed to exactly solve for the position of the Earth at each point in time as it falls to the Sun, the formula you wind up with is:

Time = 1.1 ( R^{3} / GM)^{1/2}.

So the two formulas are the same to an accuracy of about 10 percent, which is pretty good! The real answer is about 65 days, but the back of the envelope answer is more than good enough to say "about two months," like I did at the top of this essay.

*Caveats:*: Some readers will probably have objections to the above discussion and the back of the envelope method in general. Here are some questions that might be raised:

**Does the back of the envelope method always work?** No, in problems that are a lot more complicated than this one, there might be more than one way to put together all your variables to get the right units for your answer. It works here because the problem is simple enough (with only three variables: **M**, **R** and **G**) that there is really no other possible way the variables could be arranged in the final formula. If you have some physics background, there are other ways to do a back of the envelope calculation that work in a wider variety of situations. In the problem discussed here, for example, you could use conservation of energy at two different points on the Earth's trip to the Sun to get an estimate for the correct formula. Try this and see what you get! *Answer:* If you choose the starting position and the "halfway point" as your two locations, then conservation of energy reduces to the equation GM/R = V^{2}/2, where V is the speed of the Earth at the halfway point. Of course, we know that the speed of the Earth will increase as it falls, so it is hard to figure out exactly what it is at any given point, but if you estimate V as "R divided by time," then you get a final formula of:

Time = 0.7 ( R^{3} / GM)^{1/2},

which, to an order of magnitude, is the same formula as above. Using it will give you a *less* accurate answer in this case (a little over 1 month as opposed to 2 months), but it is still valuable for an order of magnitude estimate - the calculation is definitely good enough to tell you that the Earth won't crash into the sun within mere hours, nor will it wait hundreds of years!

For information on other ways to calculate the answer to this problem (including the full method with integral calculus, as well as a neat method involving orbital physics), see the PhysLink answer to this question. (The method involving orbital physics is a nice shortcut that works here but is not generally applicable to other problems in the same way that the order of magnitude estimate is. Basically, if you view the Earth falling into the Sun as one half of a very elongated elliptical orbit with an axis length equal to the Earth-Sun distance, then you can use Kepler's laws to show that the time must equal 0.5^{5/2} years, which is about 65 days.)

**Why don't we have to worry about other quantities besides M, R and G?** Off the top of your head, you might think that some other things, such as the mass of the Earth, the size of the Earth and the size of the Sun, could influence the answer to the above question. However, if we look up the numbers, we can see that these three quantities are relatively negligible. The mass of the Earth is around 5.97 x 10^{27} grams, which is a few hundred thousand times less massive than the Sun quoted above! So its gravity is certainly going to be negligible compared to the Sun's gravity in this problem (and, if you know a little physics, the rate at which the Earth falls to the Sun under the influence of the Sun's gravity *alone* is not going to depend on the Earth's mass, for the same reason that a feather and hammer will fall at the same rate in the absence of air resistance).

Similarly, the radius of the Earth is around 6.37 x 10^{8} centimeters, and the radius of the Sun is around 6.96 x 10^{10} centimeters. These are, respectively, around 24,000 and 215 times smaller than the Earth-Sun distance quoted above! So, for example, while you would be technically correct in arguing that it takes a shorter amount of time for the Earth to fall to the Sun's *surface* than it would take to fall to the Sun's center under the assumption that all the Sun's mass is located there (which is technically what we were calculating above), it turns out that the difference is negligibly small. Not only is the "extra distance" only 1/215th the size of the total distance, but this is further compounded by the fact that the Earth picks up speed as it falls. By the time it reaches the surface of the Sun, the Earth is moving very fast, so the "extra time" it would take to fall to the Sun's center is *extremely* short (on the order of minutes) when compared to the total time it takes to fall (on the order of a couple months, as we saw above).

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