Question

a) The comet Encke has an aphelion distance of 6 . 1 × 10^11 m. and perihelion distance of 5.1×10^11 .the mass of the sun is 2×10^30kg. Find the speed of the comet at the perihelion and the aphelion.


b)The planet earth is 1 . 5 × 10 11 m from the sun and orbits the sun in one year. The planet Pluto takes 248 years to orbit the sun. How far is Pluto from the sun?

Answer 1

Answer on Question#40141, Physics, Astronomy | Astrophysics

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

Solution:

a) Given:

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

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

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

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

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

ME = KE + PE = constant

$a =$ distance at aphelion

$b =$ distance at perihelion

$$\mathrm {M E} = \frac {m v ^ {2}}{2} - m \frac {G M}{r} = c o n s t a n t$$

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

When $\mathrm{PE} > \mathrm{KE}$ the comet will have an elliptical orbit with its total mechanical energy given by

ME = -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

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

b) The planet Earth is $1.5 \times 10^{\wedge} 11 \, \text{m}$ from the sun and orbits the sun in one year. The planet Pluto takes 248 years to orbit the sun. How far is Pluto from the sun?

Solution:

The third Kepler law captures the relationship between the distance of planets from the Sun, and their orbital periods.

"The square of the orbital period of a planet is directly proportional to the cube of the semimajor axis of its orbit."

Mathematically, the law says that the expression $P^2 / a^3$ has the same value for all the planets in the solar system, where $P$ is period and $a$ is distance from the Sun.

$$\frac {P _ {E a r t h} ^ {2}}{a _ {E a r t h} ^ {3}} = \frac {P _ {P l u t o} ^ {2}}{a _ {P l u t o} ^ {3}}$$ $$a _ {P l u t o} = a _ {E a r t h} ^ {3} \sqrt {\frac {P _ {P l u t o} ^ {2}}{P _ {E a r t h} ^ {2}}}$$ $$a _ {P l u t o} = 1. 5 \cdot 1 0 ^ {1 1} \sqrt {\frac {2 4 8 ^ {2}}{1 ^ {2}}} = 1. 5 \cdot 1 0 ^ {1 1} \cdot 3 9. 4 7 3 = 5 9. 2 1 \cdot 1 0 ^ {1 1} \mathrm {m}$$

Answer. a) $\mathsf{v}_{\mathrm{a}} = 14.1 \, \mathrm{km/s}$ , b) $\mathsf{v}_{\mathrm{b}} = 16.9 \, \mathrm{km/s}$ ;

b) $59.21 \times 10^{11} \, \text{m}$ .