The distance of Venus from Sun is $1.082 \times 10^{11}$ m and its period of revolution is 244.633 days. If the period of revolution of the Earth around the Sun is 365.247 days, calculate its distance from the sun.

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Solution

Kepler's Third Law of Planetary Motion. If the orbital period is given in years and the semi-major axis of the orbit is given in AU, then the period squared is equal to the semi-major axis cubed.
therefore,
$p^{2} = a u^{3} p = \sqrt{{0.73}^{3}} p = 0.62 y e a r s$

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The distance of Venus from Sun is 1.082×1011m and its period of revolution is 244.633 days. If the period of revolution of the Earth around the Sun is 365.247 days, calculate its distance from the sun.

Kepler's Third Law of Planetary Motion. If the orbital period is given in years and the semi-major axis of the orbit is given in AU, then the period squared is equal to the semi-major axis cubed. therefore, p2=au3p=√0.733p=0.62 years

The distance of Venus from Sun is $1.082 \times 10^{11}$ m and its period of revolution is 244.633 days. If the period of revolution of the Earth around the Sun is 365.247 days, calculate its distance from the sun.

Kepler's Third Law of Planetary Motion. If the orbital period is given in years and the semi-major axis of the orbit is given in AU, then the period squared is equal to the semi-major axis cubed.
therefore,
$p^{2} = a u^{3} p = \sqrt{{0.73}^{3}} p = 0.62 y e a r s$