If an object fell toward the Sun from the Earth's distance, with no orbital motion, how long would it take...?

Question

to reach the Sun?

Answer 1

answer: 64.6 days (changed later to 64.560 days)

This is a little tricker than I first thought. Intuitively, I figured the answer would be something of the order of half a year.

The potential energy of an object of mass m in the gravitational field of a large body of mass M is
U = -(GMm)/r
where r = distance between bodies.

M is the mass of the sun, m the mass of the small object.

Let
R1 = distance from sun to earth
R2 = radius of sun

We can use conservation of energy and the definition of kinetic energy to say the following:

mv^2 /2 - (GMm/r) = -(GMm / R1)

Therefore, as the object falls from a distance R1 from the sun to R2, it has velocity
v = sqrt (2 ((GM/r) - (GM / R1)))

So we know the velocity v as a function of r, the distance from the sun. We need to integrate this to get the position as a function of time, and then calculate the answer.

Unfortunately, I'm rusty at integration, so I wrote a little program to do a numerical integration instead. (For the very earliest part of the trip to the sun, I simply assumed the normal physics of a falling object in a constant gravitational field. This was necessary because the initial velocity is 0, so we have to "kickstart" the integration from a point where the velocity is non-zero.)

The result of this -- and the answer to your question -- is
64.6 days.

I think I did this correctly, but would be interested in hearing any other answers, particularly from people who see how to do the integration analytically.

-- added a few minutes later:
It's easy to calculate an upper limit to this time. Simply assume that the gravitational force between here and the sun matches that found at earth's distance from the sun. (The gravitational force actually increases as we approach the sun; that's why this calculation gives an upper limit.)

We first calculate the acceleration of gravity from the sun at the earth's distance:
a = GM / R1^2

We then use the standard formula for distance versus time in a constant gravitational field
s = (1/2) * a * t^2
to calculate the time. The result is 82.2 days. Thus, the correct answer *must* be less than 82.2 days. That gives me more confidence that the number I gave above (64.6 days) is correct.

------
Here's one more tip that the answer above is good: If you project the earth's orbit into a single axis, it makes the trip from a far point to the sun in a quarter of the year, or 91.31 days. The ratio of this time to the answer given above is 1.41. This is the square root of two, and this number shows up in physics when we talk about escape velocity versus circular velocity. So now I'm very sure that the above answer is right, but would still be interested in seeing an analytical solution from someone who's better at differential equations.

-- final edit:
I found a web page that solves this problem analytically:
http://www.physlink.com/Education/AskExperts/ae226.cfm

That page says that the answer is a quarter of a year divided by the square root of two. That quantity works out to 64.57 days, which matches the answer I found from numerical integration. (But this is a slightly incorrect approximation for reasons given below.)

Keith P's answer above is an elegant easy way of solving the problem (but, in fairness, he gave a grossly incorrect answer, and then posted the correct answer long after I posted mine). Also, technically speaking, he's solving a slightly different problem. You asked for the time to reach the sun, and I assumed you meant the surface, not the center. According to my numerical integration, the object would fall past the surface of the sun at least 12.6 minutes before it reaches the center, so the correct answer is the following:
T <= (0.5 * (0.5)^(1.5)) years - 13 minutes
so
T <= 64.56 days

The reason this is an upper limit is that I assumed the gravitational potential of a point source. Because the sun is an extended body, the acceleration of the particle through the sun (if you imagine a particle that could survive the heat) will be less than was assumed in my integration, so the travel time from surface to center will be somewhat greater than 13 minutes. I don't feel like doing this calculation now.

-- and finally, the exact approach
The web site above gives the following relation between time and distance from sun:
r*R*sqrt(1/r - 1/R) + (R^(3/2))*arctan(sqrt(R/r -1)) = sqrt(2GM)*t
where
r = distance from sun (which we set to the solar radius)
R = distance of earth from sun (1 AU)
M = mass of sun
G = gravitational constant
t = time

Plug in the numbers, and you get
t = 64.560 days
and that's my final answer! It is about 0.01 days less than Keith P's answer because it is the time to fall to the surface of the real sun, not the center of a hypothetical point sun. His approach wins the "most elegant" award, but the ugly formula above wins the "most precise" award.

In summary: The time to the surface is a bit less than Keith P's answer. The time to the center is a bit more than his answer because once the object is inside the sun, you can no longer treat the sun as a point object of fixed mass. The differences are very small, of the order of 0.01 days.

Answer 2

this is well thought, but you might also think to bring in an example of planets that are closer to the star which have acceptable temperatures, not too hot, due to a proper atmosphere, and planets that are further away that have acceptable temperatures, not too cold, due to a proper atmosphere. i know you can't, but that doesn't mean that they don't exist, and in time, i am sure that we might find one or two planets out there of such.................but, for the time being, consider that it is a perfect distance, and that the perfect distance follows within a range of 90 million to 95 million miles, that this entire range is a perfect range, that maybe if it were 89 million miles, the atmosphere couldn't support life and as for that 6 and a half mile road, it might seem like alot of distance, but when we are talking of the distance of pluto, COULD earth support life if it were as far away from the sun as pluto? If the answer is no, then, there is a range that is acceptable, and a range that is NO acceptable, despite how huge that space might seem to us, it is only a large space if we don't consider the bigger picture, like, even, outside the solar system i think this is what the above idiots were trying to get at, yet failing to do such because they tend to not think but never fear, i am sure in time, we will find planets that go against the norm...............we found a planet orbitting its star backward just recently, we are bound to find a planet that can sustain HUMAN life, but it is way too far away from its star, and we will find some explanation to it..........but, for the time being, this is what we have

Answer 3

My first ignorant guess was: "A quarter of a year. 91 days or so." But I forgot that a quarter of a plunge period isn't the same duration as a quarter of the period of the circular orbit having a radius equal to the plunge's aphelion. (Man that was dumb.)

Some of these other folks clued me in that I'd made a mistake. So here's my improved answer.

A little ball made out of something that won't melt at 5770K is dropped from a distance of one astronomical unit, or 1.4959787E+11 meters, from the center of the sun, in such a way that it has no initial speed with respect to the center of the sun.

I assumed this gravitational parameter for the sun:

GM = 1.32712440018E+20 m^3/sec^2

The ball will reach the photosphere (R=6.959512E+8 meters) after falling for 5578001 seconds, or 64.5602 days, at which time it will be moving with an inward radial velocity of 616.114 km/sec. These numbers are based on a numerical integration in 16-digit precision with a time step of 0.1 seconds.

When the time-step was increased to 1 second, the estimated time to intercept the photosphere became 5578003 sec, an increase of only 2 seconds, which indicates that the 0.1 sec time-step is satisfactory and that no further reduction in the increment is needed.

Also, the Vis Viva equation returns a speed of 616.125 km/sec at photosphere interception, a difference of only 11 m/s from the result of the integration (dt=0.1 sec). This also indicates that the integrated result is good enough.

Dr. Bob's answer, below mine, appeared before my improved answer did.

Answer 4

64.57 days.

Assume that we're thinking of a comet with an aphelion of 1 AU and a perihelion of zero. The semi-major axis is then .5 AU, and the orbital period is then (using Kepler's law) sqrt[(.5)^3] years = .35355 years. But we want only half of that: the half from aphelion to perihelion, which is .17677 years or 64.57 days.

Answer 5

I don't have a useful answer to contribute, but I'd like to point out how amazing it is that Dr. Bob and Keith P approached the same problem from completely different angles, yet arrived at the same conclusion. Well done, gentlemen! This is why I love science!

Answer 6

i do no. But , i do know , i thinks , if , the SUN WANT TO COME AND SMASH OUR EARTH A-PARTS ( or , to pieces ) , IT'S GONNA TAKE A TIMES OF ABOUTS ONLY 3 TO 4 HOURS.

Answer 7

speed of light 8eight minutes--how fast your rocket going-hummm--lets say 40,000 miles per hour-divided into 93 million miles

Answer 8

203.5 days.