Question

Two newly discovered planets follow circular orbits around a star in a distant part of the galaxy. The orbital speeds of the planets are determined to be 43.3 km/s and 58.6 km/s. The slower planet’s orbital period is 7.60 years. (a) What is the mass of the star? (b) What is the orbital period of the faster planet, in years?

Answer 1

Answer:

(a)$M = 1.62x10^{31}Kg$ (b)$T = 1.06y$

Explanation:

(a) What is the mass of the star?

The Universal law of gravitation shows the interaction of gravity between two bodies:

$F = G\frac{Mm}{r^{2}}$ (1)

Where G is the gravitational constant, M and m are the masses of the two objects and r is the distance between them.

For this particular case M is the mass of the star and m is the mass of the planet. Since it is a circular motion the centripetal acceleration will be:

$a = \frac{v^{2}}{r}$ (2)

Then Newton's second law ($F = ma$) will be replaced in equation (1):

$ma = G\frac{Mm}{r^{2}}$

By replacing (2) in equation (1) it is gotten:

$m\frac{v^{2}}{r} = G\frac{Mm}{r^{2}}$  (3)

Therefore, the mass of the star can be determine if M is isolated from equation (3):

$M = \frac{rv^{2}}{G}$  (4)

But r can be known from Kepler's third law, since it gave the semi-major axis:

$a^{3} = T^{2}$

$a = \sqrt[3]{(7.60)^{2}}$

$a = \sqrt[3]{57.76}$

$a = 3.86y$

However, a can be expressed in astronomical units:

$3.86y. \frac{1AU}{1y}$ ⇒ $3.86AU$

One astronomical unit is defined as the distance between the Earth and the Sun ($1.50x10^8 Km$):

$3.86AU. \frac{1.50x10^8km}{1AU}$ ⇒ $5.79x10^{8}Km$

r and v will be expressed in meters before being replaced in equation (4):

$r = 5.79x10^{8}Km . \frac{1000m}{1Km}$ ⇒ $5.79x10^{11}m$

$v = 43.3Km . \frac{1000m}{1Km}$ ⇒ $43300m$

$M = \frac{(5.79x10^{11}m)(43300m)^{2}}{6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2}}$

$M = 1.62x10^{31}Kg$

So the mass of the star is $1.62x10^{31}$Kg

(b) What is the orbital period of the faster planet, in years?

To find the period, the equation for orbital velocity can be used:

$v = \frac{2\pi r}{T}$  (5)

Notice that the distance of the faster planet from the Star (r) is needed, that can be found using equation (4) in terms of r and the mass of the star:

$r = \frac{MG}{v^{2}}$

It is necessary to express the velocity of the faster planet in meters.

$58.6Km . \frac{1000m}{1Km}$ ⇒ $58600m$

$r = \frac{(1.62x10^{31}Kg)(6.67x10^{-11}kg.m/s^{2}.m^{2}/kg^{2})}{(58600m/s)^{2}}$

$r = 3.14x10^{11}m$

Equation (5) can be rewritten in terms of T:

$T = \frac{2\pi r}{v}$

$T = \frac{2\pi(3.14x10^{11}m)}{58600m/s}$

$T = 33667579 s$

There are 31536000 seconds in 1 year:

$T = 33667579 s. \frac{1y}{31536000s}$ ⇒ $1.06y$

$T = 1.06y$

So the period of the faster planet is 1.06 years