 # Gravity calculations I should be able to do

I’d like to know the answer to these questions but don’t know how to work them out. (No, not homework, at my age; just curious.)

According to this site the sun’s gravitational acceleration is 274.13 m/s^s. Given that,

1. If one was in direct free fall from earth’s orbital radius to the sun, how long would it take to hit the sun and how fast would you be going when you hit it?

2. If one was, say, just inside the the farthest distance that the sun’s gravity “reaches” (does gravity “reach”?), how long would it take to reach the sun and how fast would you be going when you hit it?

Apologies for my lack of rudimentary algebra skills.

I can’t help you much myself although there are numerous people here who can.

One thing you need to bear in mind is that the acceleration figure decreases in proportion to the distance from the center of the body (IIRC in inverse proportion to the square of the distance, but I’m not sure about that). So the sunward gravitation at the distance of Earth’s position is less than it would be at the sun’s “surface”.

Right, so do you know calculus? You will need to use calculus in order to answer this problem. It’s not hard, maybe freshman-year college-level, but you need some tools first.

1. Gravity doesn’t stop “reaching”. It continues on forever as the inverse square of the distance. There is no sudden drop-off.

ETA: for 1) you could alternatively use conservation of energy, if you can look up online the potential energy of an object at earth-distance away from the sun, and then use KE=PE to solve for v

I’m relieved that it’s not as straightforward as I thought and that I should be apologizing for my lack of calculus skilz rather than algebra skilz! And in light of iamnotbatman’s correction to my incorrect concept of “reaching”, I’ll amend 2) to read:

1. If one was, say, just inside the the farthest distance that the sun’s gravity is not overcome by some other body’s gravity how long would it take to reach the sun and how fast would you be going when you hit it?

Good try, but that’s gonna be many many light years away. Someone is going to say that Proxima Centauri is the closest star, at 4.24 light years, but I’ll answer that you can just go off in a different direction away from the sun. This morphs the conversation into an astrometric question about how far one can get from the sun before he’s forced to start getting close to another star. And that’s clearly not what the OP is looking for. So I propose just picking an arbitrary point, like the distance of Jupiter or Pluto.

Alright, alright. I’ll amend 2) to:

1. If one was at the orbital radius of Sedna* (which apparently is 130 billion km from the sun at it’s most distant) how long would it take to reach the sun and how fast would you be going when you hit it?

*I only know Sedna exists b/c I googled “most distant object orbiting the sun”, just in case you were wondering why someone who knows what Sedna is couldn’t figure out the math for themselves

The calculation you can do using conservation of energy and the conversion of potential to kinetic energy. Knowing the formula for gravitational potential energy means the integration (calculus) has been done for you.

PE = GmM_sun/r_sun
KE = 1/2mv^2

PE = KE implies that:
v = sqrt(2GM_sun/r_sun) = sqrt(26.710^-112.010^30/(7.0*10^8))

give 6.2*10^5 m/s, or 1.4 million miles per hour

That’s for an object coming out from infinity. If you want to know the velocity if it just started out at earth’s orbit, you modify the above to:

PE = GmM_sun*(1/r_sun-1/d_earth)
KE = 1/2mv^2

PE = KE implies that:
v = sqrt(2GM_sun*(1/r_sun-1/d_earth)) = sqrt(26.710^-112.010^30*(1/(7.010^8)-1/(1.510^11)))

gives the same answer, given the number of significant digits I used. Basically, the object will only gain 95 thousand mph on the way to earth, and gain the rest of the 1.4 million mph on the way from the earth to the sun.

Actually, you can get a pretty good estimate to the “how long” part of the questions without calculus if you know Kepler’s Third Law. If the Sun was a perfect point mass, then you could have an orbit that had its aphelion at Earth’s orbit and its perihelion infinitesimally close to the Sun. The semi-major axis of that orbit would be half of the semi-major axis of Earth’s orbit. By Kepler’s Third Law, then, the period of such an orbit would be 1/√8 times the period of Earth’s orbit. The amount of time to fall in (and not whip around and come back out) would then be half of this. In other words, 1/(2*√8) of a year, which works out to 129 days, 3 hours, and 11 minutes.

Of course, the Sun isn’t a perfect point mass, so in real life your trip would end before you got to your “perihelion”. But I’ll wager I’m not off by more than a day.

The earth has an average distance from the Sun of 1 au, and takes 3 months to go a quarter of an orbit.

The duration of an orbit around the Sun varies as the 3/2 power of its semimajor axis length

An “orbit” that drops directly from the Earth towards the Sun has a semimajor axis of 1 au, and the trip from 1 au to the Sun is one quarter of an entire “orbit” (from 1 au to the Sun, out to 1 au, back to the Sun, and out to the original location again)

So if I’m not mistaken, the time it would take to drop directly into the Sun from 1 au is 3 months.

But I might be too sleepy at the moment to be making sense.

Oh I forgot about the question of time. Yeah, there you do need calculus.

v(t) = dr/dt, so delta_t = integral(dr/v(t))

so the time it takes to get from the earth to the sun under free fall should be:

1/sqrt(2GM)(2/3(r_earth((3/2)-r_sun^(3/2))
=1/sqrt(26.710^-112.010^30)(2/3)((1.510^11)^1.5-(710^8)*1.5)

I get 0.9 months

Correction - the “orbit” that drops directly from the Earth to the Sun has a semimajor axis of 0.5, and the trip from the Earth to the Sun is .5 of the trip. So the time it takes to drop into the Sun should be (.5)^(3/2) * .6 months or about 2.1 months. I don’t know why I’m getting a different answer than Iamnotbatman, though.

At least we have order-of-magnitude agreement! That’s good enough for me.

ETA: I’m pretty sure your calculation is an over-estimation because it doesn’t take into account the changing gravitational field as the object gets closer to the sun…

Could be, but I’d expect Kepler’s law to work pretty effectively even as you get closer to the Sun

This site http://curious.astro.cornell.edu/question.php?number=674 analyzes this problem in a couple of ways (using both my method and yours) and comes up with 2 months as an answer. I’ll think more on this after work.

Yeah, I forgot to divide by two as well. I get about 64 days, 14 hours instead.

There is no end to the reach. But at some point, you’re going to be in a position where another astronomical body will pull you more strongly than the sun does. Unless you are trying to figure out what would happen if it was just you and the sun in the entire universe…

As a point of reference, the force of gravity falls off like the square of the distance. IOW, if you are 10x the distance from earth orbit to the sun, gravity is 100x weaker.

Mike S: I’m sorry I didn’t see your post last night; we’re clearly using the same method, but you were there first.

Thank you, guys! I’ve enjoyed following along. I’m surprised it’s as complicated as it is.

I see this answer of 2 months on the internet, too. I’m a bit skeptical that if you just take the earth’s 12 month orbit and divide by 2*pi in order to get the time it would take to get to the sun at it’s current velocity, you get nearly 2 months. Maybe a coincidence. Anyways, if anyone finds the error in my calculation, I’d be curious.

No coincidence, a limiting case of Kepler’s laws. The earth’s orbit is close to a circle. the distance around the circle is 2pi*radius, the distance straight in is clearly the radius. I’ll try to check your math later.

It’s not one year divided by 2pi ≈ 6.28, it’s one year divided by 2√8 ≈ 5.66 (for the reasons explained by Andy L and I above.) The two numbers just happen to be within about 10% of each other.

I think your error’s in the integral, by the way. It looks to me like you integrated sqrt(r_earth) - sqrt® rather than 1/sqrt(1/r_earth - 1/r). (Note that these expressions are not the same.)