Question

What would be the slope of the graph of $\log r \to \log T$ obtained from the orbital from the orbital radius $r$ and the corresponding period $T$ of different satellites revolving around a planet?
A. $\dfrac{3}{2}$
B. $3$
C. $\dfrac{2}{3}$
D. $2$

Answer 1

Hint: In order to solve this question we need to understand Kepler's law of planetary motion. There are three rules of Kepler’s motion, first law states that every planet move around sun in an elliptical orbit with sun as one of the foci, second law states that a radius vector joining any planet to sun sweeps out equal area in equal interval of time, so when the planet is farthest from the sun it moves with less speed while when the planet is very close to sun then it moves with less speed and the third law states that square of the time period of revolution of planet is directly proportional to cubes of their mean distance from the sun.

Complete step by step answer:
Let the time period of revolution be, $T$ and also let the orbital radius be $r$.
So according to Kepler’s third law: ${T^2} \propto {r^3}$
Removing proportionality we get, ${T^2} = k{r^3}$
Using log on both sides we get, $\log {T^2} = \log (k{r^3})$
Using log formula we get, $2\log T = \log (k) + \log ({r^3})$
$2\log T = \log k + 3\log (r)$
$\Rightarrow 3\log r = 2\log T - \log k$
$\Rightarrow \log r = \dfrac{2}{3}\log T - \dfrac{1}{3}\log k \to (i)$
Comparing this equation with standard straight line equation: $y = mx + c$
We get, $m = \dfrac{2}{3}$
Since the slope is $m$ so the slope of the curve between $\log r$ and $\log T$ is, $\dfrac{2}{3}$.

So the correct option is C.

Note:It should be remembered that Kepler’s first law states that planets moves around sun in elliptical orbit and sun is at one of the foci, means that the distance between planet and sun is constantly changing, so the speed of planet is not same at every point in the orbit. Also the point of nearest approach to sun is called perihelion and greatest separation is known as aphelion.