Question

(a) Explain Kepler's laws of planetary motion.
(b). Deduce Newton's law of gravitation from these laws.
c) The distances of two planets from the sun are 10 raised to the power 11 m and 10 raised to the power 10 m respectively. What
time period of these two planets?

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Answer 1

Answer:

Answer ( a)

Kepler's law of planetary motion -

Kepler's First Law

It states that the path of any planet in an orbit around the sun follows the shape of an ellipse with the sun at one of the foci.

Kepler's Second Law

It states that an imaginary path coming out from the sun the planet sweeps out in equal area in equal intervals of time.

Kepler's Third Law

It states that the cube of the mean distance of a planet from the sun is directly proportional to the square of its orbital period T

Answer (2)

1) A force acting on the planet due to sun is the centripetal force which is directed towards the sun.

2) The force acting on the planet must be inversely proportional to the square of the distance from the sun.

3) The force acting on the planet is directly proportional to the product of the masses of the planet and the sun.

Answer 2

(a)

Kepler's laws of planetary motion are the Law of orbits, the law of equal areas and the law of periods.

Explanation:

• The three laws of planetary motion or Kepler's laws of planetary motion are:

    Kepler's first law of motion (Law of orbits):

• According to Kepler’s first law, "All the planets revolve around the sun in elliptical orbits having the sun at one of the foci."
• For an ellipse, the sum of the distances of any planet from two foci is constant.
• The occurrence of seasons is due to the elliptical orbit of a planet.

   Kepler's second law of motion (Law of equal areas):

• According to Kepler’s second law, "The radius vector drawn from the sun to the planet sweeps out equal areas in equal intervals of time."
• In equal intervals of time, the area swept is constant.
• Thus, Kepler’s second law can also be stated as “The areal velocity of a planet which is revolving around the sun in elliptical orbit remains constant which shows that the angular momentum of a planet remains constant”.

   Kepler's third law of motion (Law of periods):

• According to Kepler’s third law of motion, "The square of the time period of revolution of a planet around the sun in an elliptical orbit is directly proportional to the cube of its semi-major axis."
• It is represented as:

        $\[{T^2} \propto {a^3}\]$

(b)

Given:

Kepler's laws of planetary motion

To Derive: Newton's law of gravitation

Derivation:

Suppose a planet of mass $m$ is revolving around the Sun of mass $M$.

For simplification, assume that the orbits are circular.

Now, if the planet is revolving in a circular orbit, a centripetal force acts on it. This force is given as:

$\[{F_{centripetal}} = \frac{{m{v^2}}}{r}\]$                ... (i)

where, $v=$ speed of the planet, $r=$ radius of the orbit

The time period of the revolution is given as:

$\[T = \frac{{2\pi r}}{v}\]$

Thus, velocity of the planet is given as:

$\[v = \frac{{2\pi R}}{T}\]$

Substituting the value of velocity in the equation (i), we have,

$\[\begin{gathered} {F_{centripetal}} = \frac{m}{r}{\left( {\frac{{2\pi r}}{T}} \right)^2} \hfill \\ {F_{centripetal}} = \frac{{4{\pi ^2}m{r^2}}}{{r{T^2}}} \hfill \\ {F_{centripetal}} = \frac{{4{\pi ^2}mr}}{{{T^2}}} \hfill \\ \end{gathered} \]$

From Kepler's third law, we have,

$\[{T^2} = k{r^2}\]$

Now, the above equation becomes

$\[\begin{gathered} {F_{centripetal}} = \frac{{4{\pi ^2}mr}}{{k{r^2}}} \hfill \\ {F_{centripetal}} = \left( {\frac{{4{\pi ^2}}}{k}} \right)\frac{m}{{{r^2}}} \hfill \\ \end{gathered} \]$

Here, $\[{\frac{{4{\pi ^2}}}{k}}\]=constant$

Analysing the first and the second laws of planetary motion, we have,

$\[\frac{{4{\pi ^2}}}{k} = GM\]$

Where, $G=$ Universal gravitational constant and $M=$ mass of the sun

Therefore,

$\[{F_{centripetal}} = \frac{{GMm}}{{{r^2}}}\]$

This is Newton's law of gravitation from these laws.

(c)

(There is something missing in the question. The correct question is read as: The distances of two planets from the sun are 10 raised to the power 11 m and 10 raised to the power 10 m respectively. What is the ratio of the time period of these two planets?)

Given:

Distance of the first planet from the sun $\[ = {10^{11}}\,m\]$

Distance of the second planet from the sun $\[ = {10^{10}}\,m\]$

To Find: Ratio of the time period of these two planets

Solution:

According to Kepler's third law,

$\[{T^2} \propto {r^3}\]$

Therefore, the ratio of the time periods of the two planets can be given as:

$\[\begin{gathered} \frac{{{T_1}}}{{{T_2}}} = {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^{3/2}} \hfill \\ \frac{{{T_1}}}{{{T_2}}} = {\left( {\frac{{{{10}^{11}}}}{{{{10}^{10}}}}} \right)^{3/2}} \hfill \\ \frac{{{T_1}}}{{{T_2}}} = 10\sqrt {10} \hfill \\ \end{gathered} \]$

Hence, the ratio of the time period of these two planets is $\[10\sqrt {10} \]$.

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