
Two planets at mean distance ${d_1}$ and ${d_2}$ from the sun and their frequencies are ${n_1}$ and ${n_2}$ respectively then
(a) ${n_1}^2{d_1}^2 = {n_2}^2{d_2}^2$
(b) ${n_2}^2{d_2}^3 = {n_1}^2{d_1}^3$
(c) ${n_1}{d_1}^2 = {n_2}{d_2}^2$
(d) ${n_1}^2{d_1} = {n_2}^2{d_2}$
Answer
483.3k+ views
Hint: These concepts are based on Kepler’s law of planetary motion. Kepler’s third law states that the square of the period of time of the earth is directly proportional to the cube of the semi - major axis of its orbit.
Formula Used: ${T^2} \propto {r^3}$
Complete answer:
Mathematically the Kepler’s third law of planetary motion is given as,
${T^2} \propto {r^3}$, here r = d
We know that, $T = \dfrac{1}{n}$
Where ‘n’ is the frequency.
Now, from Kepler’s third law of planetary motion we get,
$\eqalign{
& \dfrac{1}{{{n^2}}} \propto {d^3} \cr
& \Rightarrow \dfrac{{{n_2}^2}}{{{n_1}^2}} = \dfrac{{{d_1}^3}}{{{d_2}^3}} \cr
& \Rightarrow {n_2}^2{d_2}^3 = {n_1}^2{d_1}^3 \cr} $
Thus, two planets at the mean distance ‘${d_1}$’ and ‘${d_2}$’ from the sun and their frequencies are ‘${n_1}$’ and ‘${n_2}$’ respectively then ${n_2}^2{d_2}^3 = {n_1}^2{d_1}^3$.
Hence, option (B) is the correct answer.
Additional Information:
The eccentricity of an elliptical orbit is defined as the measure of the amount by which it deviates from a circle. It is solved by dividing the distance between the focal points of the ellipse by the length of the major axis.
Perigee is known as the orbit of an artificial satellite or natural satellite at which the centre of the satellite is nearest to the centre of the planet. Similarly, Apogee is known as the orbit of an artificial satellite or natural satellite at which the centre of the satellite is farthest to the centre of the planet.
Note:
Kepler’s first law states that all the planets revolve around the sun in elliptical orbits with the sun at one focus. Kepler’s second law states that the line which joins the earth or any other planet to the sun sweeps out equal areas in equal intervals of time.
Formula Used: ${T^2} \propto {r^3}$
Complete answer:
Mathematically the Kepler’s third law of planetary motion is given as,
${T^2} \propto {r^3}$, here r = d
We know that, $T = \dfrac{1}{n}$
Where ‘n’ is the frequency.
Now, from Kepler’s third law of planetary motion we get,
$\eqalign{
& \dfrac{1}{{{n^2}}} \propto {d^3} \cr
& \Rightarrow \dfrac{{{n_2}^2}}{{{n_1}^2}} = \dfrac{{{d_1}^3}}{{{d_2}^3}} \cr
& \Rightarrow {n_2}^2{d_2}^3 = {n_1}^2{d_1}^3 \cr} $
Thus, two planets at the mean distance ‘${d_1}$’ and ‘${d_2}$’ from the sun and their frequencies are ‘${n_1}$’ and ‘${n_2}$’ respectively then ${n_2}^2{d_2}^3 = {n_1}^2{d_1}^3$.
Hence, option (B) is the correct answer.
Additional Information:
The eccentricity of an elliptical orbit is defined as the measure of the amount by which it deviates from a circle. It is solved by dividing the distance between the focal points of the ellipse by the length of the major axis.
Perigee is known as the orbit of an artificial satellite or natural satellite at which the centre of the satellite is nearest to the centre of the planet. Similarly, Apogee is known as the orbit of an artificial satellite or natural satellite at which the centre of the satellite is farthest to the centre of the planet.
Note:
Kepler’s first law states that all the planets revolve around the sun in elliptical orbits with the sun at one focus. Kepler’s second law states that the line which joins the earth or any other planet to the sun sweeps out equal areas in equal intervals of time.
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