Two very small planets orbit a much larger star. Planet A orbits the star with a period TA . Planet B orbits the star at four times the distance of planet A, but planet B is four times as massive as planet A. Assume that the orbits of both planets are approximately circular.

Planet B orbits the star with a period TB equal to

a. TA/4.
b. 16TA.
c. 8TA.
d. 4TA
d. TA

Explanation:

Given that,

Period of Planet A around the star

Pa = Ta

Let the semi major axis of planet A from the star be

a(a) = x

Given that, the distance of Planet B from the star (i.e. it semi major axis) is 4 time that of Planet A

a(b) = 4 × a(a) = 4 × x

a(b) = 4x

Also planet B is 4 times massive as planet A

Using Kepler's law

P² ∝a³

P²/a³ = k

Where

P is the period

a is the semi major axis

Then,

Pa² / a(a)³ = Pb² / a(b)³

Pa denotes Period of Planet A and it is Pa = Ta

a(a) denoted semi major axis of planet A and it is a(a) = x

Pb is period of planet B, which is the required question

a(b) is the semimajor axis of planet B and it is a(b) = 4x

So,

Ta² / x³ = Pb² / (4x)³

Ta² / x³ = Pb² / 64x³

Cross multiply

Ta² × 64x³ = Pb² × x³

Divide both sides by x³

Ta² × 64 = Pb²

Then, Pb = √(Ta² × 64)

Pb = 8Ta

Then, the period of Planet B is eight times the period of Planet A.

The correct answer is C

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s