**Question 11.**

**Solve the following problems.If the height of a satellite completing one revolution around the earth in T seconds is h1 meter, then what would be the height of a satellite taking 2√2 T seconds for one revolution?**

Answer:

Given:

The height of the satellite completing one revolution around the earth in T seconds.

let the height be x of a satellite taking 2√2 T seconds for one revolution.

From Kelper’s third law of motion:

T ∝ r3/2

WhereT is the time period

r is the radius of the satellite

Thus we can say,

T = K r3/2 ....(i)

Here, r1 = r when, T1 = t

We get,

Then, if T2 = 2√2t

Putting the value in the equation(i), we get

2T = K r13/2

Putting the value of K in the above equation, we get

On solving the above equation, we get

⇒

Squaring both side and then taking cube-root

.....(ii)

so using these relations we get

x=R + 2h1