Observations with the Hubble Space Telescope revealed a planet about 16 AU from a star whose mass is the same as our Sun’s mass. How long does it take the planet to orbit the star?

Since the star's mass equals that of the sun, you can use the same form of Kepler's third law that applies to our solar system.

P^2 = a^3

In the above formula, P is the period in years and a is the distance from the star in a.u.

See http://spiff.rit.edu/classes/phys440/lectures/kepler3/kepler3.html
for more about the law

P^2 = (16)^3 = 4096
P = 64 years

To determine the orbital period of a planet, we can use Kepler's third law of planetary motion, which states that the square of a planet's orbital period is directly proportional to the cube of its average distance from the star.

In this case, we know that the planet is located about 16 Astronomical Units (AU) from its star, which has the same mass as our Sun.

To calculate the orbital period, we can use the following equation:

T^2 = (4π^2 / GM) * r^3

Where:
T is the orbital period in seconds.
G is the gravitational constant (6.67430 × 10^-11 m^3 kg^-1 s^-2).
M is the mass of the star, which is equal to the solar mass (1.989 × 10^30 kg).
r is the distance between the star and the planet, expressed in meters.

First, we need to convert the distance from AU to meters. Since 1 AU is approximately 149,597,870,700 meters, the distance in meters would be:

r = 16 AU * 149,597,870,700 meters / AU = 2.39357 × 10^12 meters.

Now we can substitute the values into the formula:

T^2 = (4π^2 / (6.67430 × 10^-11 m^3 kg^-1 s^-2 * 1.989 × 10^30 kg)) * (2.39357 × 10^12 meters)^3.

Simplifying the equation gives us:

T^2 ≈ 1.449054 × 10^25 seconds^2.

To find the orbital period T, we take the square root of both sides:

T ≈ √(1.449054 × 10^25 seconds^2).

Calculating this, we find:

T ≈ 3.80576 × 10^12 seconds.

However, to express this time in a more sensible unit, we can convert seconds to years. There are approximately 31,536,000 seconds in a year, so we can divide the calculated time by this value:

T ≈ (3.80576 × 10^12 seconds) / (31,536,000 seconds/year) = 1.20778 × 10^5 years.

Therefore, it would take the planet roughly 120,778 years to complete one orbit around its star.