Question

The planet Jupiter has an elliptical orbit with e = 0.05 and a semi-major axis of
8.7 10 m.
11 × Calculate the energy of the planet, perihelion and aphelion distances and the
speed of the planet at these points.

Answer 1

Answer on Question #54941-Physics-Mechanics-Kinematics-Dynamics

The planet Jupiter has an elliptical orbit with $e = 0.05$ and a semi-major axis of $7.8 \cdot 10^{11} \, \text{m}$ .

Calculate the energy of the planet, perihelion and aphelion distances and the speed of the planet at these points.

Solution

The energy of the planet is

$$E = \epsilon \frac {m M}{m + M},$$

where $\epsilon$ is specific orbital energy, $m$ is mass of Jupiter, $M$ is the mass of Sun.

$$\epsilon = - \frac {G (m + M)}{2 a}.$$ $$E = - \frac {G (m + M)}{2 a} \frac {m M}{m + M} = - \frac {G (m + M)}{2 a} = - \frac {6 . 6 7 \cdot 1 0 ^ {- 1 1} (1 . 9 8 8 \cdot 1 0 ^ {3 0} + 1 . 8 9 8 \cdot 1 0 ^ {2 7})}{2 \cdot 7 . 8 \cdot 1 0 ^ {1 1}} = - 8. 5 \cdot 1 0 ^ {7} J.$$

Aphelion distance is

$$r _ {a} = (1 + e) a = 8. 2 \cdot 1 0 ^ {1 1} m.$$

Perihelion distance is

$$r _ {p} = (1 - e) a = 7. 4 \cdot 1 0 ^ {1 1} m.$$

Under standard assumptions the orbital speed $v$ of a body traveling along an elliptic orbit can be computed from the Vis-viva equation as:

$$v = \sqrt {\mu \left(\frac {2}{r} - \frac {1}{a}\right)}.$$

$\mu$ is the standard gravitational parameter. For Jupiter

$$\mu = 1 2 6 6 8 6 5 3 4 \frac {k m ^ {3}}{s ^ {2}} = 1 2 6 6 8 6 5 3 4 \cdot 1 0 ^ {9} \frac {m ^ {3}}{s ^ {2}}.$$

For aphelion

$$v = \sqrt {1 2 6 6 8 6 5 3 4 \cdot 1 0 ^ {9} \left(\frac {2}{8 . 2 \cdot 1 0 ^ {1 1}} - \frac {1}{7 . 8 \cdot 1 0 ^ {1 1}}\right)} = 3 8 3 \frac {m}{s}.$$

For perihelion

$$v = \sqrt {1 2 6 6 8 6 5 3 4 \cdot 1 0 ^ {9} \left(\frac {2}{7 . 4 \cdot 1 0 ^ {1 1}} - \frac {1}{7 . 8 \cdot 1 0 ^ {1 1}}\right)} = 4 2 4 \frac {m}{s}.$$

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