(a) At what height above Earth’s surface is the energy required to lift a satellite to that height equal to the kinetic energy required for the satellite to be in orbit at that height? (b) For relater heights, which is greater, the energy for lifting or the kinetic energy for orbiting?

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.