19.0k views
0 votes
A newly discovered planet orbits a distant star with the same mass as the Sun at an average distance of 112 million kilometers. Its orbital eccentricity is 0.3. Find the planet’s orbital period and its nearest and farthest orbital distances from its star.

User FLOZz
by
6.8k points

1 Answer

6 votes
To find the orbital period of a planet, we can use the equation:

T = 2 * pi * (a^3 / (G * M))^(1/2)

where T is the orbital period, a is the semi-major axis (average distance) of the orbit, G is the gravitational constant, and M is the mass of the star.

First, we'll convert the average distance from kilometers to meters:

a = 112 million km = 112 x 10^6 m

Next, we'll find the semi-minor axis using the formula:

b = a * sqrt(1 - e^2)

where e is the eccentricity.

b = 112 x 10^6 m * sqrt(1 - 0.3^2) = 112 x 10^6 m * sqrt(0.91) = 112 x 10^6 m * 0.954

Next, we'll find the semi-major axis:

a = 112 x 10^6 m

Finally, we'll substitute these values into the formula for the orbital period:

T = 2 * pi * (a^3 / (G * M))^(1/2)

T = 2 * pi * (112 x 10^6 m)^3 / (G * M)^(1/2)

Since M is the mass of the sun, we'll use the mass of the sun in kilograms:

M = 1.989 x 10^30 kg

G = 6.67430 x 10^-11 m^3 kg^-1 s^-2

T = 2 * pi * (112 x 10^6 m)^3 / (6.67430 x 10^-11 m^3 kg^-1 s^-2 * 1.989 x 10^30 kg)^(1/2)

T = 2 * pi * (112 x 10^6 m)^3 / (6.67430 x 10^-11 * 1.989 x 10^30)^(1/2)

T = 2 * pi * (112 x 10^6 m)^3 / (1.327 x 10^-20)^(1/2)

T = 2 * pi * (112 x 10^6 m)^3 / (1.153 x 10^-10)

T = 2 * pi * (112 x 10^6 m)^3 / 1.153 x 10^-10

T = 2 * pi * (112 x 10^6 m)^3 / 1.153 x 10^-10

T = 2 * pi * (112 x 10^6 m)^3 / 1.153 x 10^-10

T = 5.201 x 10^7 s

So, the orbital period of the planet is approximately 5.201 x 10^7 seconds, or about 1.6 years.

The nearest and farthest orbital distances from the star are equal to the semi-minor and semi-major axes, respectively.

Nearest distance = b = 112 x 10^6 m * 0.954 = 106.6 x 10^6 m

Farthest distance = a = 112 x 10^6 m
User Billjk
by
6.6k points