The mass of mars is 6.38x10^23 kg and its radius is 3.38 x10^6m. mars rotates on its axis with a period of 1.026 days.(g=6.67x10^-11 nm^2/kg^2). calculate the orbital speed for a satellite at an altitude of 1.62x10^6 m.

v = 2917.35 m/s

Explanation:

let Fc be the centripetal force avting on the satelite , Fg is the gravitational force between mars and the satelite, m is the mass of the satelite and M is the mass of mars.

at any point in the orbit  the forces acting on the satelite are balanced such that:

Fc = Fg

mv^2/r = GmM/r^2

v^2 = GM/r

   v = \sqrt{GM/r}  

      = \sqrt{(6.6708×10^-11)(6.38×10^23)/(3.38×10^6 + 1.62×10^6)}

      = 2917.35 m/s

Therefore, the orbital velocity of the satelite orbiting mars is 2917.35 m/s.

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .