# 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...
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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

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:
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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...
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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/p...

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

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/p...

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)...
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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

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

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