A planet is discovered orbiting the star 51 Peg with a period of four days (0.01 years). 51 Peg has the same mass as the Sun. Mercury's orbital period is 0.24 years, and Venus's is 0.62 years.

The average orbital radius of this planet is:

a) less than Mercury's.

b) between Mercury's and Venus's.

c) greater than Venus's.

Question

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Physics

9 January, 16:20

A planet is discovered orbiting the star 51 Peg with a period of four days (0.01 years). 51 Peg has the same mass as the Sun. Mercury's orbital period is 0.24 years, and Venus's is 0.62 years.

The average orbital radius of this planet is:

a) less than Mercury's.

b) between Mercury's and Venus's.

c) greater than Venus's.

+2

Answers (1)

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Cordell Acevedo · Answer 1

a) less than Mercury's.

Explanation:

For orbital time period of a planet, the expression is

T² = 4π² R³ / GM

T is time period, R is radius of orbit, G is universal gravitational constant, M is mass of the star or sun

T² ∝ R³

As radius of orbit increases, time period increases. The given planet is making around a star whose mass is equal that of sun so M is same as sun.

The given planet has time period equal to. 01 years which is less than that of Mercury and Venus, hence its R will be less than orbit of both of them or less than mercury's.