Question

The comet Encke has an aphelion distance of 6.1× 10^11m and perihelion distance of
5.1× 10^11m.The mass of the sun is 2.0 ×10^30kg.Find the speed of the comet at the
perihelion and the aphelion.

Answer 1

Answer on Question #42081, Physics, Molecular Physics | Thermodynamics

The comet Encke has an aphelion distance of $6.1 \times 10^{11} \mathrm{~m}$ and perihelion distance of $5.1 \times 10^{\wedge 11} \mathrm{~m}$ . The mass of the sun is $2.0 \times 10^{30} \mathrm{~kg}$ . Find the speed of the comet at the perihelion and the aphelion.

Solution:

a) Given:

$a =$ aphelion $= 6.1\times 10^{11}\mathrm{m}$

$\mathsf{b} =$ perihelion $= 5.1\times 10^{11}\mathrm{m}$

$M = 2.0 \times 10^{30} \mathrm{~kg}$

$\mathsf{v_a} = ?,\mathsf{v_b} = ?$

The total mechanical energy (TE) of a comet, or any orbiting body, is the sum of its kinetic energy (KE) and its gravitational potential energy (PE):

TE = KE + PE = constant

TE = $\frac{mv^2}{2} - m\frac{GM}{r} = \text{constant}$

where $M$ is the mass of the Sun, $m$ is the mass of the comet, $r$ is its instantaneous distance from the Sun and $G$ ( $6.673 \times 10^{-11} \, \text{Nm}^2 \, \text{kg}^{-2}$ ) is the universal gravitational constant.

When PE > KE the comet will have an elliptical orbit with its total mechanical energy given by

TE = -m $\frac{GM}{2s}$

$-m\frac{GM}{2s} = \frac{mv^2}{2} -m\frac{GM}{r}$

The comet has a velocity given by

$v = \sqrt{\frac{GM}{s}\left(\frac{2s}{r} - 1\right)}$

where $s$ is the mean radius of its orbit (sometimes referred to as the semi-major axis).

The aphelion + perihelion = the major axis.

$s = \frac{a + b}{2} = 5.6\times 10^{11}\mathrm{m}$

The speed of the comet at the aphelion

$$v _ {a} = \sqrt {\frac {G M}{s} \left(\frac {2 s}{a} - 1\right)} = \sqrt {\frac {6 . 6 7 3 \cdot 1 0 ^ {- 1 1} \cdot 2 \cdot 1 0 ^ {3 0}}{5 . 6 \cdot 1 0 ^ {1 1}} \left(\frac {2 \cdot 5 . 6}{6 . 1} - 1\right)} = 1 4 1 1 5. 7 \frac {\mathrm {m}}{\mathrm {s}} = 1 4. 1 \mathrm {k m / s}$$

The speed of the comet at the perihelion

$$v _ {b} = \sqrt {\frac {G M}{s} \left(\frac {2 s}{b} - 1\right)} = \sqrt {\frac {6 . 6 7 3 \cdot 1 0 ^ {- 1 1} \cdot 2 \cdot 1 0 ^ {3 0}}{5 . 6 \cdot 1 0 ^ {1 1}} \left(\frac {2 \cdot 5 . 6}{5 . 1} - 1\right)} = 1 6 8 8 3. 5 \frac {\mathrm {m}}{\mathrm {s}} = 1 6. 9 \mathrm {k m / s}$$

Answer. $\mathsf{v}_{\mathsf{a}} = 14.1 \mathsf{km} / \mathsf{s},$

$$v _ {b} = 1 6. 9 \mathrm {k m / s}.$$

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