Question

Kepler's third law states that square of period of revolution (T) of a planet around the sun, is proportional to third power of average distance r between sun and planet i.e. $T^2 = Kr^3$ here K is constant. If the masses of sun and planet are M and m respectively then as per Newton's law of gravitation force of attraction between them is $F = \frac {GMm}{r^2}$, here G is gravitational constant. The relation between G and K is described as

• A. $K = G$
• B. $K = \frac {1}{G}$
• C. $GK = 4 \pi^2$
• D. $GMK = 4 \pi^2$

Answer 1

Answer: (D)

$GMK = 4 \pi^2$

Answer 2

Gravitational force of attraction between sun and planet provides centripetal force for the orbit of planet.
$\therefore \frac {GMm}{r^2} = \frac {mv^2}{r}$
v^2 = \frac {GM}{r} \hspace20mm ... (i)
Time period of the planet is given by
$T=\frac{2\pi r}{v}$, $T^{2}=\frac{4\pi^{2}r^{2}}{v^{2}}$
$T^2 = \frac {4\pi^2r^3 }{\bigg(\frac{GM}{r} \bigg)}$ [Using equation (i)]

T^2 = \frac {4\pi^2r^3}{GM} \hspace20mm ... (ii)
According to question,
T^2 = Kr^3 \hspace20mm ... (iii)
Comparing equations (ii) and (iii), we get
$K= \frac {4\pi^2}{GM} \therefore GMK = 4\pi^2$