# Answer to Question #11506 in Mechanics | Relativity for Nero

Question #11506

A satellite rotates around mars at 1 revolution in 1.96 hours. How high above the mars surface is the satellite?

Expert's answer

A satellite rotates around mars at 1 revolution in 1.96 hours. How high above the mars surface is the satellite?

Using

a = GM/R²,

where R is the radius from the center of the Mars, M is the mass of the Mars and G is the Gravitational constant and also use the centripetal force equation:

a = W²R,

where W is the radians / second velocity and R is the radius from the center of the Mars. Using 1 revolution in 1.96 hours, you can compute the W as 2π/(3600*1.96).

Combining these equations and cancelling the a (acceleration) gives

GM/R² = W²R.

Solving for R we get

R³ = GM/W² ==> R = (CM/W²)^(1/3).

If you plug in all the numbers you will end up with a radius R which can then be subtracted from the Mars's radius to obtain the height above the Mars's surface.

Using

a = GM/R²,

where R is the radius from the center of the Mars, M is the mass of the Mars and G is the Gravitational constant and also use the centripetal force equation:

a = W²R,

where W is the radians / second velocity and R is the radius from the center of the Mars. Using 1 revolution in 1.96 hours, you can compute the W as 2π/(3600*1.96).

Combining these equations and cancelling the a (acceleration) gives

GM/R² = W²R.

Solving for R we get

R³ = GM/W² ==> R = (CM/W²)^(1/3).

If you plug in all the numbers you will end up with a radius R which can then be subtracted from the Mars's radius to obtain the height above the Mars's surface.

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