
How is orbital period calculated if perihelion and aphelion are known? For example, the orbit of a spacecraft about the sun has a perihelion distance of 0.5AU and aphelion of 3.5AU. What is its orbital period?
Answer
528.3k+ views
Hint: You could find the orbital period from by using kepler’s third law. You may not find the perihelion and aphelion in the common expression but their sum is the semi major axis which would be the R in the formula. Since all we have is proportionality relation, we could find their ratio which would be a constant by substituting earth’s semi major axis and time period and then find the orbital period for the given spacecraft by substituting accordingly.
Formula used:
Kepler’s third law of planetary motion,
${{T}^{2}}\propto {{R}^{3}}$
Complete answer:
We could find the orbital period by using Kepler’s third law of planetary motion when the perihelion and aphelion are known.
According to Kepler’s third law,
${{T}^{2}}\propto {{R}^{3}}$
Where, R is the body’s average distance from the sun and T is the time period.
$\dfrac{{{T}^{2}}}{{{R}^{3}}}=C$ ………………………………….. (1)
Now let us find this constant C by substituting Earth’s orbital period and semi major axis.
$C=\dfrac{{{\left( 1Year \right)}^{2}}}{{{\left( 1AU \right)}^{3}}}=1\dfrac{yea{{r}^{2}}}{A{{U}^{3}}}$ ………………………………….. (2)
Now, for the spacecraft,
$R=\dfrac{perihelion+aphelion}{2}$
$\Rightarrow R=\dfrac{5AU+3.5AU}{2}=2AU$ ………………………………. (3)
Now from equation (1) we have,
${{T}^{2}}=C{{R}^{3}}$
Substituting (2) and (3) we get,
${{T}^{2}}=\left( 1\dfrac{Yea{{r}^{2}}}{A{{U}^{3}}} \right){{\left( 2AU \right)}^{3}}$
$\Rightarrow {{T}^{2}}=8year{{s}^{2}}$
$\therefore T=2\sqrt{2}years$
Hence, we found the spacecraft’s orbital speed to be 2.83years.
Note:
Kepler’s third law is not limited to the sun and its planetary system. It could be applied to all orbiting bodies. However, the constant that we found would quite obviously be different and hence should be found accordingly. Also, AU stands for astronomical unit which is used as a unit for measuring greater distances.
Formula used:
Kepler’s third law of planetary motion,
${{T}^{2}}\propto {{R}^{3}}$
Complete answer:
We could find the orbital period by using Kepler’s third law of planetary motion when the perihelion and aphelion are known.
According to Kepler’s third law,
${{T}^{2}}\propto {{R}^{3}}$
Where, R is the body’s average distance from the sun and T is the time period.
$\dfrac{{{T}^{2}}}{{{R}^{3}}}=C$ ………………………………….. (1)
Now let us find this constant C by substituting Earth’s orbital period and semi major axis.
$C=\dfrac{{{\left( 1Year \right)}^{2}}}{{{\left( 1AU \right)}^{3}}}=1\dfrac{yea{{r}^{2}}}{A{{U}^{3}}}$ ………………………………….. (2)
Now, for the spacecraft,
$R=\dfrac{perihelion+aphelion}{2}$
$\Rightarrow R=\dfrac{5AU+3.5AU}{2}=2AU$ ………………………………. (3)
Now from equation (1) we have,
${{T}^{2}}=C{{R}^{3}}$
Substituting (2) and (3) we get,
${{T}^{2}}=\left( 1\dfrac{Yea{{r}^{2}}}{A{{U}^{3}}} \right){{\left( 2AU \right)}^{3}}$
$\Rightarrow {{T}^{2}}=8year{{s}^{2}}$
$\therefore T=2\sqrt{2}years$
Hence, we found the spacecraft’s orbital speed to be 2.83years.
Note:
Kepler’s third law is not limited to the sun and its planetary system. It could be applied to all orbiting bodies. However, the constant that we found would quite obviously be different and hence should be found accordingly. Also, AU stands for astronomical unit which is used as a unit for measuring greater distances.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Trending doubts
10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Explain zero factorial class 11 maths CBSE

What is a periderm How does periderm formation take class 11 biology CBSE

