If the distance between sun and earth were reduced to half its present value, then the number of days in one year would have been
129
Kepler's Law of Planetary Motion
To solve this problem, we need to apply Kepler's Third Law of Planetary Motion, also known as the Law of Periods. This law describes the relationship between the orbital period of a planet and the semi-major axis (average distance) of its orbit around the Sun.
Kepler's Third Law states that the square of the orbital period (\(T\)) of a planet is directly proportional to the cube of the semi-major axis (\(R\)) of its orbit. Mathematically, this can be expressed as:
\[T^2 \propto R^3\]
This proportionality can also be written as:
\[\frac{T^2}{R^3} = \text{constant}\]
This means that for any two objects orbiting the same central body (like planets orbiting the Sun), the ratio of the square of their periods to the cube of their average distances will be the same.
Applying Kepler's Third Law for Reduced Distance
Let's denote the initial conditions (present value) with subscript '1' and the new conditions (reduced distance) with subscript '2'.
- Present orbital period of Earth (\(T_1\)) = 365 days (approximately).
- Present distance between Sun and Earth (\(R_1\)).
- New distance between Sun and Earth (\(R_2\)) = \(\frac{R_1}{2}\) (reduced to half its present value).
- We need to find the new number of days in one year (\(T_2\)).
According to Kepler's Third Law:
\[\frac{T_1^2}{R_1^3} = \frac{T_2^2}{R_2^3}\]
Now, we can substitute the given values into the equation:
\[T_2^2 = T_1^2 \left( \frac{R_2}{R_1} \right)^3\]
Substitute \(T_1 = 365\) days and \(R_2 = \frac{R_1}{2}\):
\[T_2^2 = (365)^2 \left( \frac{\frac{R_1}{2}}{R_1} \right)^3\]
Simplify the ratio of distances:
\[T_2^2 = (365)^2 \left( \frac{1}{2} \right)^3\]
Calculate the cube of \(\frac{1}{2}\):
\[T_2^2 = (365)^2 \left( \frac{1}{8} \right)\]
To find \(T_2\), take the square root of both sides:
\[T_2 = \sqrt{\frac{(365)^2}{8}}\]
\[T_2 = \frac{365}{\sqrt{8}}\]
We know that \(\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}\).
\[T_2 = \frac{365}{2\sqrt{2}}\]
To rationalize the denominator, multiply the numerator and denominator by \(\sqrt{2}\):
\[T_2 = \frac{365 \times \sqrt{2}}{2\sqrt{2} \times \sqrt{2}}\]
\[T_2 = \frac{365 \times \sqrt{2}}{2 \times 2}\]
\[T_2 = \frac{365 \times \sqrt{2}}{4}\]
Using the approximate value \(\sqrt{2} \approx 1.414\):
\[T_2 = \frac{365 \times 1.414}{4}\]
\[T_2 = \frac{516.11}{4}\]
\[T_2 \approx 129.0275\]
Therefore, if the distance between the Sun and Earth were reduced to half its present value, the number of days in one year would be approximately 129 days.
Summary of Calculation
| Parameter | Current Value | New Value |
|---|---|---|
| Orbital Period (\(T\)) | \(T_1 = 365\) days | \(T_2 = ?\) |
| Distance (\(R\)) | \(R_1\) | \(R_2 = R_1 / 2\) |
Using the relationship \(T^2 \propto R^3\), we calculated \(T_2\) to be approximately 129 days.
Important Questions from Kepler’s laws
Which of the following laws represented by the formula T2/R3, compares the orbital period and radius of the orbit of a planet with that of other planets?
A planet is at distance 'a' from the sun and its time period of revolution is T yrs. The planet suddenly cames to distance \(\rm\frac{a}{2}\) closer to the sun. The new time period of revolution is:
Kepler’s law of “Areal Velocity is constant” is equivalent to law of conservation of
If the earth is 1/4 of its present distance from the sun, then the duration of the year would be
Two planets orbit the Sun in circular orbits, with their radius of orbit as R 1= R and R 2 = 4R. Ratio of their periods (T 1/T 2) around the Sun will be
