Respuesta :
Since the orbit is circular and you are given the period and average distance( in other words the radius of the orbit) we can calculate the orbital speed right away. Keep in mind that ship's mass is much smaller than the mass of the moon, this is really important, as we can simply say that moon is not affected by the ship.
Angular frequency and orbital speed have the following relationship:
[tex]v=wr=\frac{2\pi}{T}\cdot r[/tex]
[tex]v=\frac{2\pi}{133\cdot 60}1.99345\cdot 10^6=1569.5795\frac{m}{s}[/tex]
Let's do the calculation with rest of the numbers just to make sure.
Because you have the stable circular orbit the gravitational force and centrifugal force are in balance:
[tex]G\frac{mM}{r^2}=m\frac{v^2}{r}[/tex]
We solve this for v:
[tex]G\frac{M}{r}=v^2\\ v=\sqrt{G\frac{M}{r}}[/tex]
You can use this formula for any celestial body as long as the ship's mass is much smaller than the mass of the body, and this only applies to circular orbits
When we plug in the numbers we get:
[tex]v=1569.5799\frac{m}{s}[/tex]
As you can see the results are identical.
Angular frequency and orbital speed have the following relationship:
[tex]v=wr=\frac{2\pi}{T}\cdot r[/tex]
[tex]v=\frac{2\pi}{133\cdot 60}1.99345\cdot 10^6=1569.5795\frac{m}{s}[/tex]
Let's do the calculation with rest of the numbers just to make sure.
Because you have the stable circular orbit the gravitational force and centrifugal force are in balance:
[tex]G\frac{mM}{r^2}=m\frac{v^2}{r}[/tex]
We solve this for v:
[tex]G\frac{M}{r}=v^2\\ v=\sqrt{G\frac{M}{r}}[/tex]
You can use this formula for any celestial body as long as the ship's mass is much smaller than the mass of the body, and this only applies to circular orbits
When we plug in the numbers we get:
[tex]v=1569.5799\frac{m}{s}[/tex]
As you can see the results are identical.