How do I calculate the mass and orbital distance of a planetary body?

[revolv]

Hey everyone. I am working on a planetary simulation model and am having a bit of trouble with the math. Given the volume and density of a particular planetary body, how do I calculate the mass? Once I have the mass, how do I calculate the orbital distance of a satellite body to said panetary body? I'm sure this is somewhat elementary astrophysics but, in the immortal edited words of Dr. McCoy, I'm a programmer, not a physicist! Thanks in advance for any help.

 

[Andrew Mason]

Hey everyone. I am working on a planetary simulation model and am having a bit of trouble with the math. Given the volume and density of a particular planetary body, how do I calculate the mass? Once I have the mass, how do I calculate the orbital distance of a satellite body to said panetary body? I'm sure this is somewhat elementary astrophysics but, in the immortal edited words of Dr. McCoy, I'm a programmer, not a physicist! Thanks in advance for any help.

Density is mass/volume:\rho = m/V so m = \rho V

The orbital radius of a satellite (distance from satellite to centre of the planet) will depend on the speed of the satellite. A satellite can orbit at any distance.

AM

 

[revolv]

Thanks for the info, Andrew. It is very much appreciated. One other question if I may. If i know the distance, what is the equation for determining the orbital speed? Or, from a education standpoint, where can i find said equation?

 

[Andrew Mason]

Thanks for the info. It is very much appreciated. One other question if I may. If i know the distance, what is the equation for determining the orbital speed? Or, from a education standpoint, where can i find said equation?

The force of gravity provides the centripetal acceleration, so:

mv^2/r = GMm/r^2 so:

v = \sqrt{GM/r} where G is the universal gravitation constant and M is the mass of the planet

G = 6.67 x 10^-11 m³/kg sec²

Example: The mass of the Earth is M = 5.98 x 10²⁴kg

The radius of the Earth at the equator is about 6,378,000 m. So, at a distance of 1000 km above the earth, the orbital speed would be:

v = \sqrt{6.67e-11*5.98e24/7.378e6} = 7.35 x 10^3 m/sec

This works out to about 26,500 km/hr.

AM

 

[revolv]

Thanks a bunch, Andrew. That is exactly wahat I needed.

-B.