A probe is fired directly away from the sun. How long until it hits the sun?

The probe is initial 1 au from the sun. Its initial velocity is 1 au/year. This velocity is directly away from the sun in the reference frame of the sun. (i.e., in a direction radial to the sun).

The probe will travel directly outwards, and then it will slow down, stop and begin falling directly towards the sun until it hits it.

This is solvable without calculus by using Kepler's equations

Update:

*************

I originally intended that this problem be the one that Kirchwey solved. Noting that if the probe had 2π au/year velocity, and was at 1au from earth, it would have the same semi-major axis as earth and have a 1 year degenerate orbit. The problem would then be fairly easy to solve:

http://www.cramster.com/answers-feb-10/physics/gra...

Giving an answer of (3/2π + 1) / 2π = 0.909 years

Oh, well, I miswrote the question. Let's figure how Al, Dr D, and Lithium stand up.

.........

The new major axis will be:

2a = 1 + 1/(8 π^2) ≈ 1.01266 au

and the semi-major axis will be

a = 0.50633 au

If we now expand the ellipse with out loss of generality, the areas are much easier to calculate. (see, above citation for a how to). The area swept out in the up direction to reach aphelion will be:

φ = acos (r.e-a)/a

φ = acos (1 - 0.50633)/0.50633 au

φ = 0.224 rads

Update 2:

A = πa^2 φ/2π + a^2/2 *sin φ )

A = a^2 ( φ + sinφ /2)

The area down will of course be:

A.d = π/2 a^2

The period of orbit will be

P = (a/r.e)^3/2

P = (0.50633)^3/2

P = 0.3602 years

The time 'til collision with sun center:

T = P * (A.up +A.down) /( π a^2)

T = P * ( a^2 ( φ + sinφ /2) +π/2 a^2 ) /( π a^2)

T = 0.2186 years

T = 79.8 days

8 Answers

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  • 10 years ago
    Favorite Answer

    I'd rather start with the question you said you intended to ask in the previous posting, where initial velocity is 2pi au/yr. Even with this simplification I still couldn't get it all from Kepler.

    With this velocity the probe has the same KE, PE and total E as the 1 au circular orbit.

    PE0 = -GMm/r0, KE0 = -PE0/2 and E0 = PE0+KE0 = PE0/2.

    The probe is in a degenerate elliptical (i.e., linear) orbit whose max PE = E0 = PE0/2 thus apoapsis = 2r0. Of course periapsis = 0. Then the semimajor axis = r0 and T1 = T0.

    Fall time = T1/2 = 0.5 yr

    Partial-distance rise time (from r0 to 2r0) is the same as fall time over that distance, which is a complicated function of distance (ref.):

    t = sqrt(y0^3/(2*mu)) * (sqrt(y/y0*(1-y/y0)) + arccos(sqrt(y/y0))), where mu = GM, y0 is initial height = 2r0 and y is height at time t = r0.

    Solving, rise time = 0.409155 yr

    Total time to crash = 0.909155 yr or 28690158 s.

    I really don't have the time to scale all this to a different velocity. If lithiumdeuteride cares to provide his derivation it would be hugely appreciated. I can't quite make sense of his velocity figures; with a 24 s time difference at 2E6 m/s, the solar radius comes out to only 48000 km vs. actual value of ~350000 km.

    EDIT: OK, I found the time to program a calculator for this and solve for v = 1 au/yr. My answer basically agrees with LiD's.

    KE (relative to circ. orbit) is reduced by a factor of 4pi^2 thus KE = -PE0/(8pi^2).

    Adding KE to PE0 and solving,

    apoapsis = r0*1/(1-1/(8pi^2)) = r0*1.01282761155353

    SMA = apoapsis/2

    T1 = T0*(SMA/r0)^1.5

    Full-distance fall time = T1/2 = 0.5*T0*(0.5*1.01282761155353)^1.5 = 0.180189014324892 yr

    Partial-distance rise time (from r0 to apoapsis) = 2.57646416217077E-02 yr

    Total time to impact = 0.2059536559466 yr = 6499270 s

    Re: jean-de-la-lune, I think we agree that KE = -PE/2 applies to any circular orbit. In this case we have initial PE of Earth's orbit, and with 2pi au/yr we have KE of Earth's orbit, so I don't see a problem with starting the analysis from the circ. orbit. I continued with the circ. orbit basis for comparison for the 1 au/yr case because that's how I started. Lazy, I guess. But the relation between the period and the SMA is the same for any ellipse, and thus the full-distance straight-down fall time is just half the period of an ellipse with infinitesimal width.

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  • 10 years ago

    If the probe was launched directly away from the Sun, it won't hit the Sun. Instead, it's speed would just reduce. Due to the Earth's distance from the Sun, any speed directly away from the Sun is enough to escape the Solar System. Unless it passes by an object that has enough gravity to cause the probe to change course. Then it would spiral in toward the Sun. But if the probe is traveling at 10,000s miles per hour, the only type of object in the Solar System that could cause it to change course a lot would be a giant planet. However, if it was launched from the Earth and not from a spacecraft in space not drifting, it would go in a curve anyways and not directly away from the Sun, and it would eventually begin to spiral in toward the Sun.

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  • 3 years ago

    The earth is incredibly nearer to the solar in the time of wintry climate. The earth's orbit isn't changing/ The warming is as a results of having too lots greenhouse gases, and that's a sluggish substitute of a few levels on hassle-free in a hundred years. the earth won't hit the solar, however the solar will enhance right into a brilliant pink super in approximately 4 billion years. which will consume up the Earth. The Christians will burn too. bang your head against a heavy Bible, possibly which will help. possibly you're able to desire to stay at school longer.

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  • Dr D
    Lv 7
    10 years ago

    If I'm understanding the question correctly, we are ignoring the movement of the sun during this period. And we have the differential equation:

    r"(t) = -k/r^2

    where k = 4π^2 au^3 / yr^2 approximately

    subject to initial conditions

    r(0) = 1 au

    r'(0) = 1 au/yr

    And we need to solve for when r = 0.004652 au (sun radius).

    The difficulty is in solving this analytically. It can easily be solved numerically.

    I'm getting 0.2059295 years = 75.2 days

    *****

    OK, there is an analytical approach. Start with

    r^2 * r" = -k

    Write v = r', therefore

    r^2 dv/dt = r^2 dv/dr * dr/dt = -k

    r^2 * v dv/dr = -k

    v * dv = -k/r^2 * dr

    Integrating with the initial conditions

    v^2 - 1 = 2k * (1/r - 1)

    v^2 = 1-2k + 2k/r

    dr/dt = √ (1-2k + 2k/r)

    dt = ± dr / √ (1-2k + 2k/r)

    This is +ve during the ascent, and -ve during the descent.

    This expression can be integrated, although it is really difficult to type out here.

    dr/dt = 0 when r = 2k/(2k-1)

    We can then find the time taken to reach the peak, the time taken to descend, then add them up.

    I'm getting 0.2059298179 years or 75.21586599 days

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  • 10 years ago

    If the Sun were a point mass, impact would be at approximately:

    t = 6,498,620 s

    But since it's not, impact instead occurs at approximately:

    t = 6,498,596 s

    At the time of impact, the probe is traveling at a speed of approximately:

    v = 1,956,930 m/s

    This solution doesn't take special relativity into account, but since the final impact velocity is only 0.006528 times light speed, the error should be small.

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  • 10 years ago

    This is a most confusing question. I withdrew my answer to the first version because I felt dissatisfied with the extrapolation to an 'ellipse' which would be a segment. And I think it is wrong and you have no right to apply Kepler's laws to such a trajectory. Mathematically, an ellipse turns into a parabola when eccentricity -->1 or it can degenerate into a pair of straight lines when you write the general conic section equation and impose certain conditions on the coefficients. But a segment, I don't see how. Physically, a trajectory through the center of force is completely different from a trajectory which has angular momentum.

    Kirchway's equality KE = -PE/2 holds only because he modifies the assignement and takes 2Pi au/y velocity, which allows him to apply results valid for the earth. But of course KE=-PE/2 does not hold generally for a trajectory which ends in a finite time (the demonstration of the virial theorem does not hold) It does not hold if you take 1 au/y as initial velocity of your degenerate trajectory.

    Flamechop and Magnetulsar are completely wrong since even if you had postulated a 2Pi au/y velocity, it would still be short by a factor sqrt(2) from the escape velocity. (and I assume 8th grade astronomy class is a joke)

    So your answer can be found by integrating 1/sqrt[2(GM/r+C)] with C = v_0^2/2 - GM/Re first from Re to Rmax (with Rmax from C = -GM/Rmax) and then from Rmax to 0, but I don't see any way to derive it using Kepler's laws. Even if you did use them for the segment 0-Rmax seen as an ellipse of semi major axis Rmax/2, (which is not warranted in my opinion but it might be that the integral gives you the right relation for the 'half year'-you'll only know it by integrating !) you would still have to calculate the time from Re up to Rmax using integration.

    Edited: I have just skimmed through dear old Lev Davidovitch and in (very) short: conservation of angular momentum (L) is equivalent to the area law which by integration yields 2m_eS = Te L with S the orbit area. Knowing it is an ellipse and having related the semi axes to G, M, L, E and m_e by using the conservation laws to solve for theta(r), you can eliminate both S and L (and m_e) and get Kepler's third law. Of course this one no longer depends on finite surface or non zero angular momentum, but you simply cannot derive it without using them.

    To Sum Up: if you start from Newton, which is arguably the basis, there is no way to deduce Kepler's law for a zero angular momentum trajectory. (Besides, they are absolutely meaningless for such a trajectory) You have one and only one equation which is d^2r/dt^2 = -GM/r^2 which gives you a first (energy) integral, 1/2dr/dt-GM/r = C and by integrating that one can answer your question, but there is no way to short cut calculus.

    Re: kirchway, of course I agree with KE = -PE/2 (for any periodical orbit actually) so you can use earth data to calculate initial KE in terms of Rearth by scaling the speed and then find Rmax.

    My contention is that you have no right to use T^2 =kR^3 for the degenerate orbits going through the center because to demonstrate it, you need T to be a period (not a countdown time to an end) and R to be the SMA of a finite area ellipse.

    This is a question of physics, not of math. Now since there is no apparent discontinuity in the formula, you can bet it is usable for the limiting case also. However you can only be sure by explicit integration (and of course, it works)

    It remains that

    a/ even with this somewhat dubious 'Kepler's law', you still have to perform an integration for the Re-->Rmax part (Done in your complicated formula)

    b/ your final answer must be wrong..see below

    LiD's remark about relativity not being important is valid only thanks to the finite solar size. Otherwise the speed grows without limit to compensate for PE --> -infinity and keep the total energy constant and finite. Then at some point one can no longer claim v/c << 1 Because of that, LiD's time for the pointlike sun is certainly wrong and so is yours if your sun is pointlike

    Actually there is an almost consistent way to consider the degenerate trajectory as the limit of an ellipse by using the angular momentum L as a parameter -->0 Keeping the energy fixed, the eccentricity-->1 (1-e^2 ~ L^2) the major axis is fixed, the minor axis-->0 (b ~ L) and the point of closest approach ->0 as L^2

    The inconsistency lies in that this point is reached for theta=pi, that is, the 'other side' of the center with respect to what happens with the L=0 trajectory. Clearly, there can't exist a consistent way of treating angles when there are none in the limit.

    The approximate time up calculation of AI P is beautiful but evidently wouldn't work with v=2Pi au/y

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  • 10 years ago

    sorry but more information is needed for example, mass of probe and/or mass of sun in order to calculate the acceleration acting on the probe.

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  • 10 years ago

    Never, space is infinite and an object wouldn't fall back, it would keep going until an outside force was acted on it.

    Source(s): Me!, + 8th grade astronomy class!
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