Difficult falling body problem?

Question

how long , T(tot), does a mass m with initial velocity =0, take to fall to the surface of a planet (of mass M and radius r) from a height H above the surface , in a non-constant field given by GMm/(r+s)^2?

(where s is the height above the surface at any time t)

this is a lot more difficult than it looks!

full points for an elegant solution.

(full working details if you can thanks)

I thought of that too steve but for a circular orbit the acceleration is constant beacuse the height is constant.

alexander
------------------
if "Orbital period is determined by semi-major axis alone"
then PROVE

its equal to the period of circular orbit of radius ρ = (R+H)/2.

SECONDLY
IT DOES NOT AT ALL "become clear" that part of area of full ellipse swept in downfall is
S/π = 1/2π [arccos(R/ρ - 1) + R/ρ √(HR/ρ²) ]

What does S/Pi mean anyway
and where is the derivation of your answer?
Answer:
T = √(MG/ρ³) [arccos(R/ρ - 1) + √(HR³) /ρ² ]

Answer 1

Eh Steve H.,

From my experience, Alexander is almost always right (although he will make mistakes or flip variables). My understanding is that he has a doctrate in Physics. And what is clear to him may not be clear to the rest of us.

First, all ellipses and a circle that have the same semi-major axis (just axis in the case of a circle) have the same period. See http://en.wikipedia.org/wiki/Orbital_period#Small_body_orbiting_a_central_body . Using this formula, the time to hit the center of the earth (which would be a foci and 2xsemi-major axis away from the start point) would be

T=1/2 orbit period = 1/2 * (2*pi * sqr(a^3/GM)

a is the semi-major axis and is also equal to (R+h)/2

T=pi *sqr(a^3/GM)

also Tp= Time of Orbital Period = 2T

Now you need to calculate back to the earth surface R. Kepler's third law about equal areas in equal time is the way to go (Very Elegant, but I would not have realized doing it this way except for Alexander). See http://en.wikipedia.org/wiki/Orbital_mechanics#Derivation

The S/pi term is simply the area of downfall compared with the total area of the elipse. Just a proportion.

I'll trust that Alexander did the area equation correctly
S/π = 1/2π [arccos(R/a - 1) + R/a √(HR/a²) ]

therefore, the Total Time

Tt= Tp * 1/2π [arccos(R/a - 1) + R/ρ √(HR/a²) ]
= 2pi *sqr(a^3/GM) * 1/2π [arccos(R/a - 1) + R/ρ √(HR/a²) ]
= sqr(a^3/GM) * [arccos(R/a - 1) + R/ρ √(HR/a²) ]

Note: Alexander flipped the variables in the equation for time of orbit. This derivation is also dependent on Alexander's obvision sight for the area of the ellipse. It looks right to me, (I looked at R=0), but I'd have to derive it myself to be sure and frankly, that would make my head hurt.

Answer 2

I saw this problem stated in an old issue of the American Journal of Physics as, "If the earth suddenly stopped in its orbit, how long would it take to fall into the sun?"

Their elegant solution was to treat it as an orbit problem where the orbit just went back and forth through the sun to the far side and back. The claim was that the period of the orbit is the same no matter what the semimajor axis of the ellipse is, even when it is zero.

Since the earth's orbital period is a year, the trip from aphelion to the sun is 1/4th the whole period, so it would take 3 months.

If I'm remembering things correctly then you can find the time it takes to fall to your planet by just computing a quarter of the orbital period for a circular orbit at the starting height.

This actually gives the time it would take to get all the way to the center of the planet, so it will only be valid for H >> r.