Question

Astronomers can calculate the mass of a planet by using the orbital parameters of one of its moons. Consider a hypothetical planet, which has a hypothetical moon that orbits the planet in a roughly circular orbit with a radius of 3.95*10^8 m. The moon takes 12 Earth days to orbit the planet. What is the mass of the planet?

Answer 1

To find the mass of the planet, we can use the formula for the gravitational force acting between two masses:

F = G * (m1 * m2) / r^2

where F is the gravitational force, G is the gravitational constant (6.674 x 10^-11 m^3 kg^-1 s^-2), m1 and m2 are the masses of the two objects, and r is the distance between them.

We can also express the force acting on the moon as:

F = m_moon * a

where m_moon is the mass of the moon, and a is its centripetal acceleration due to its orbit around the planet.

Since both expressions describe the same force, we can set them equal to each other:

m_moon * a = G * (m_planet * m_moon) / r^2

The centripetal acceleration can be calculated using:

a = v^2 / r

where v is the orbital speed of the moon. We can find this speed using the circumference of the circular orbit and the orbital period:

v = (2 * pi * r) / T

where T is the orbital period in seconds. Since the moon takes 12 Earth days to orbit the planet:

T = 12 days * (24 hours/day) * (3600 seconds/hour) = 1,037,600 seconds

Now, we can plug this value into the equation for the orbital speed:

v = (2 * pi * 3.95 * 10^8 m) / 1,037,600 s ≈ 7,569 m/s

Then we can find the centripetal acceleration:

a = (7,569 m/s)^2 / (3.95 * 10^8 m) ≈ 1.456 m/s^2

Now, we can plug this acceleration back into the equality of forces:

m_moon * (1.456 m/s^2) = G * (m_planet * m_moon) / (3.95 * 10^8 m)^2

Notice that the mass of the moon (m_moon) cancels out on both sides of the equation. We can now solve for the mass of the planet:

m_planet = (1.456 m/s^2) * (3.95 * 10^8 m)^2 / (6.674 * 10^-11 m^3 kg^-1 s^-2) ≈ 3.29 * 10^24 kg

So the mass of the hypothetical planet is approximately 3.29 * 10^24 kg.