91Ó°ÊÓ

The value of energy required to lift the satellite and the kinetic energy required to put the satellite in the orbit needs to be the same. Therefore,$\left(\mathrm{U}\right)=\mathrm{K}$ .

The difference between Potential energy on the earth’s surface and at height h will give us $\left(U\right)$. By comparing it with thekinetic energy of the orbiting satellite at height, we can calculate (h).Using the same equation; we can decide the greater quantity between $\mathbf{\left(}\mathbf{U}\mathbf{\right)}$and K.

Formula:

$$\begin{gathered} U=U_f-U_i \\ {U}_f=\frac{-GMm}{\left(R+h\right)} \\ {U}_i=\frac{-GMm}{\left(R\right)} \\ K=\frac{GMm}{2\left(R+h\right)} \end{gathered}$$

Calculation for h:

$$\begin{gathered} U=Uf-Ui \\ U=\frac{-GMm}{\left(R+h\right)}-\frac{-GMm}{\left(R\right)} \\ K=\frac{GMm}{2\left(R+h\right)} \end{gathered}$$

(i)

Equating equation (i) and (ii)

$$\begin{gathered} \frac{1}{R}-\frac{1}{R+h}=\frac{1}{2\left(R+h\right)} \\ \frac{1}{2\left(R+h\right)}+\frac{1}{\left(R+h\right)}=\frac{1}{R} \\ \frac{1}{2\left(R+h\right)}+\frac{2}{2\left(R+h\right)}=\frac{1}{R} \\ \frac{3}{2\left(R+h\right)}=\frac{1}{R} \\ 2R+2h=3R \\ h=\frac{R}{2} \\ =\frac{6370}{2} \\ =3185km \end{gathered}$$

At $h=3185km$ from the surface of the earth, energy required for lifting the satellite is equal to the kinetic energy of the orbiting satellite

Calculation for comparing U and K:

At greater heights, let. Using this condition in the equation (i) and (ii),

$$\begin{gathered} U=\frac{-GMm}{h} \\ K=\frac{GMm}{2h} \end{gathered}$$

The denominator of is less than K by factor 2.

Hence U > K .

At greater heights, the energy required for lifting the satellite is greater than the kinetic energy of the satellite in the orbit.