find the orbital velocity of a satellite placed on the orbit of radius 9000 km. Radius of the earth = 6400km.

THE ANS IS 5157 M/S

For the satelite revolving around the earth in a circular orbit of radius R,

 where M is the mass of earth NOW, R= r + H

H = 9000 KM AND r = 6400 KM HENCE

R = 9000 +6400 =15400 km

NOW YOU CAN CALCULATE 

The method is absolutely correct.We have,

Orbital velocity V is given by V²=GM/R where R=r+h=15,400km

Putting G=6.67X10^-11 and M=6X10²⁴ S.I units,we get,

V²=6.67X10^-11X6X10²⁴/154X10⁵=(6.67X3/77)X10⁸.

Since you want the exact answer,I calculated using calculator and got the exact answer as 5097.74m/s.But my method will be like this,

6.67X3=(20/3)(3)=20 and (20/77)(10⁸)=(2000/77)(10⁶).Now,2000/7 is approximately 26.

So,V²=26X10⁶ V=5.1X10³=5100.(as rt26 is app. 5.1).

Hence,the answer I got is V=5100m/s.

u could modify the formula like this n solve it.

V²⁼ GM/R+h as R²g/R+h   and solve it.

Velocity of a satellite revolving the earth is

V_{o =}  

G=6.67 * 10^-11Nm²kg^-2     M=6.01 * 10²⁴ kg

R=6400 km = 6.4 * 10⁶ m    h=9000 km = 9 * 10⁶m

Substitute the values 

V_o=

   _{Find the value of this} 

The answer has been given by allamraju, but, in case you are interested in knowing how that formula is obtained, here goes:

The satellite as such moves around the earth in an elliptical orbit, which is assumed to be almost circular for simplicity. Now, any object moving in a circle requires that a force acts on it towards the center which keeps it moving in that circle. (For example, when you whirl a stone tied to a string, it moves in a circle because the string keeps the stone from breaking off and moving away. If the string ruptures, the stone would move away, wouldn't it?)

This central force is called the centripetal force, and its magnitude is given by , where R is the radius of the orbit, and v is the velocity of the object, Here, for the satellite, the force acting towards the center of the orbit is nothing but the gravitational force of attraction on the satellite due to the earth, and we know that its magnitude is given by .

Hence, for the satellite to move in a circular path, we must have

Which boils down to the result .

If you've understood this, even if you forget the formula, you'll be able to derive it in seconds.

In this question it is convenient to use this formula:

V_o= R