How can the orbital semi-major axis, eccentricity, and period be calculated if perihelion and aphelion are known?

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

\sqrt{C R^{2}}

\sqrt{C R^{3}}

\sqrt{C R^{4}}

\sqrt{C R^{5}}

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Concept:

To find semi-major axis:
We know that the longest diameter of the ellipse is called the major-axis. The half of the major axis is called the semi-major axis.
We can find the semi-major axis by knowing the perihelion and aphelion distance.
In the ellipse, the shortest distance from the centre of the ellipse is called perihelion. The farthest distance from the centre is called aphelion. In Greek, ‘Peri’ means close and ‘Apo’ means far. And ‘Helio’ means sun. Therefore here the ellipse is the orbit of the Earth around the sun.
The sum of perihelion and the aphelion gives the value of the major axis as the sum defines the largest diameter. By just dividing the major axis, we can find semi-major axes. Lets take R is the average distance of the Earth from the sun. The perihelion value is taken as Rp and the aphelion value is taken as Ra
Therefore,

R = \frac{(Rp + Ra)}{2}

This R value gives the value of the semi-major axis.
The value of eccentricity defines how much the ellipse is elongated. The value of eccentricity is denoted by ‘e’. The value of ‘e’ lies in between 0 and 1
The perihelion value is taken as Rp and the aphelion value is taken as Ra

\Rightarrow Ra = 1 + e

\Rightarrow Rp = 1 - e

Dividing these two values give,

\frac{Ra}{Rp} = 1 + e / 1 - e

\Rightarrow Ra (1 - e) = Rp (1 + e)

\Rightarrow Ra - Rp = (Rp + Ra) e

∴ e = \frac{(Ra - Rp)}{(Rp + Ra)}

Hence, the value of eccentricity can be found by perihelion and aphelion.
According to Kepler’s 3rd law, the square of the orbital time period is directly proportional to the cube of the average distance from the sun i.e.

T^{2} a R^{3}

∴ \frac{T^{2}}{R^{3}} = C

\Rightarrow T = \sqrt{C R^{3}}

We already found the value of R by using perihelion and aphelion values. Now we can find the orbital time period by using the value of R.
Hence, option 2 is the correct answer.

Watch 3-min video & get full concept clarity