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The following problem is giving me a headache:

Halley's comet follows an elliptical orbit around the sun. At perihelion, its distance from the sun ($r_P$) is $8.823 \cdot 10^{10}$ metres. At aphelion, its distance from the sun ($r_A$) is $6.152 \cdot 10^{12}$ metres. Its perihelion speed ($v_P$) is $5.46 \cdot 10^4$ metres per second. Compute its aphelion speed ($v_A$).

While conservation of angular momentum does give me the correct answer, conservation of mechanical energy doesn’t, and I don’t understand why.

Conservation of angular momentum:

$m v_A r_A = m v_P r_P$

$\displaystyle{v_A = \frac{r_P}{r_A} \, v_P \approx 783 \text{ m s}^{-1}}$

Conservation of mechanical energy:

$\displaystyle{\frac{1}{2} m {v_A}^2 - \frac{G m M_\odot}{r_A} = \frac{1}{2} m {v_P}^2 - \frac{G m M_\odot}{r_P}}$

$\displaystyle{v_A = \sqrt{ {v_P}^2 + 2 G M_\odot \!\left( \frac{1}{r_A} - \frac{1}{r_P} \right)} \approx 3939 \text{ m s}^{-1}}$

The following problem is giving me a headache:

Halley's comet follows an elliptical orbit around the sun. At perihelion, its distance from the sun ($r_P$) is $8.823 \cdot 10^{10}$ metres. At aphelion, its distance from the sun ($r_A$) is $6.152 \cdot 10^{12}$ metres. Its perihelion speed ($v_P$) is $5.46 \cdot 10^4$ metres per second. Compute its aphelion speed ($v_A$).

Using two methods:

1. Conservation of angular momentum:

$m v_A r_A = m v_P r_P$

$\displaystyle{v_A = \frac{r_P}{r_A} \, v_P \approx 783 \text{ m s}^{-1}}$

2. Conservation of energy:

$\displaystyle{\frac{1}{2} m {v_A}^2 - \frac{G m M_\odot}{r_A} = \frac{1}{2} m {v_P}^2 - \frac{G m M_\odot}{r_P}}$

$\displaystyle{v_A = \sqrt{ {v_P}^2 + 2 G M_\odot \!\left( \frac{1}{r_A} - \frac{1}{r_P} \right)} \approx 3939 \text{ m s}^{-1}}$

Question:

While conservation of angular momentum does give me the correct answer, conservation of mechanical energy doesn’t. Why is this so?