- #1

FranzDiCoccio

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## Homework Statement

The period of a comet is 75.8 years. The perihelion distance is 0.596 AU (1 AU = 1.5 ⋅ 10

^{11}m).

The velocity at perihelion is v

_{p}= 5.45 ⋅10

^{4}m/s.

a) Find the length of the major semi-axis of the elliptical orbit.

b) Find the aphelion distance and the velocity at aphelion for the comet.

## Homework Equations

T

^{2}= k a

^{3}(Kepler's third law)

m v r = const. (conservation of angular momentum at aphelion and perihelion)

[tex]1/2 m v^2 - \frac{G m M}{r} = const[/tex] (conservation of energy)

## The Attempt at a Solution

[/B]

I used the 3rd Kepler's law to find the semi-major axis a. For the constant I used

[tex]k=\frac{4 \pi^2}{G M}[/tex]

where M = 1.99 ⋅10

^{30}kg as provided by the book.

I find

a = 17.9 AU = 2.68⋅10

^{12}m

r

_{a}= 35.2 AU = 5.28 ⋅10

^{12}m

Then, from [tex]m v_a r_a = m v_p r_p[/tex] I find v

_{a}=923 m/s.

So far, I'm in

**complete agreement with the solution provided by the book**.

Now, I think that the same result could be derived from the conservation of energy.

[tex]v_a = \sqrt{v_p^2+2GM \left(\frac{1}{r_a}-\frac{1}{r_p}\right)}[/tex]

However, this formula gives a completely different result: 7.15 ⋅10

^{3}m/s.

Much larger than the previous one.

For the life of me, I cannot see the mistake. Here we are talking about a small comet orbiting the Sun, so the latter should be basically at rest. Center of mass should not be relevant.

The other thing I can think of is the constant in Kepler's law. But that should be ok in this case (again, the comet has a very small mass)

Can anyone help? Could it perhaps be a matter of roundoff? Difference between very small numbers?

EDIT: corrected typo in initial formula for energy

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