The moon Phobos orbits Mars (m = 6.42 x 1023 kg) at a distance of 9.38 x 106 m. What is the Phobos' orbital velocity?

Answer 1

We must take two force equations and set them equal.

The first is F = mv²/r for the centripetal motion of phobos

The second is F = Gm₁m₂/r² for the gravitational attraction between phobos and mars

Setting them equal we get:

m₁v²/r = Gm₁m₂/r²

where G = the gravitational constant, m₁ = the mass of phobos, and m₂ = the mass of mars

m₁ cancels on both sides so we are left with:

v²/r = Gm₂/r²

Solving the equation for v we get:

v = sqrt(Gm₂/r)

plugging everything in gives us:

v = sqrt(6.67E-11 * 6.42E23/ 9.38E6)

v = 2136.63 m/s

Answer 2

Hi Tamia!

Since you've asked a couple of similar questions, I'll write a general approach to this problem.

In general, when you are being asked about orbital velocity/period/radius, you are using the same basic principle. Assuming the orbit is circular, you are dealing with uniform circular motion, where the centripetal force is the gravitational force between the planet and the orbiting body. This gives you the following equation:

GMm/r² = mv²/r,

Where r is the radius between the planet and the orbiting object, M is the mass of the planet, m is the mass of the orbiting object, v is the velocity of the orbiting object, and G is the gravitational constant.

You can use this equation to solve for velocity or radius. To solve for period, you can solve for velocity first. Then note that the distance covered in one period is 2*pi*r. You then have T = 2*pi*r/v.

Hope this helps - let me know if this is unclear!