First slide

Keplers 3 Laws

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. $T^{2} = K r^{3}$ 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
$F = \frac{G M m}{r^{2}}, here G is gravitational constant.$
The relation between G and K is described as

Moderate

A

K = G
B

$K = \frac{1}{G}$
C

$G K = 4 π^{2}$
D

$G M K = 4 π^{2}$

Solution

Gravitational force of attraction between sun and planet provides centripetal force for the orbit of planet.
$∴ \frac{G M m}{r^{2}} = \frac{m v^{2}}{r}$
$v^{2} = \frac{G M}{r}$ . . . .(I)
Time period of the planet is given by
$T = \frac{2 π r}{v}, T^{2} = \frac{4 π^{2} r^{2}}{v^{2}}$
$T^{2} = \frac{4 π^{2} r^{2}}{(\frac{G M}{r})}$ (using eqn. (i))
$T^{2} = \frac{4 π^{2} r^{3}}{G M}$ . . . . (ii)
According to question
$T^{2} = K r^{3}$ . . . .(iii)
$Comparing equations (ii) and (iii), we get$
$K = \frac{4 π^{2}}{G M} ∴ G M K = 4 π^{2}$

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