By what percent the energy of a satellite has to be increased to shift it from an orbit of radius r to 3r/2

The correct option is 33.3%.

To find the percentage increase in the energy of the satellite, we first recall the formula for the total energy of a satellite orbiting a planet.

The total energy $E$ of a satellite of mass $m$ in a circular orbit of radius $r$ around a planet of mass $M$ is given by:
$E=-\frac{GMm}{2r}$
where $G$ is the universal gravitational constant.

Let $E_1$ be the initial energy of the satellite in the orbit of radius $r$:
$E_1=-\frac{GMm}{2r}$

Let $E_2$ be the final energy of the satellite in the orbit of radius $r_2=\frac{3r}{2}$:
$E_2=-\frac{GMm}{2\left(\frac{3r}{2}\right)}=-\frac{GMm}{3r}$

The change in energy ($\Delta E$) required to shift the satellite is:
$\Delta E=E_2-E_1$
$\Delta E=-\frac{GMm}{3r}-\left(-\frac{GMm}{2r}\right)$
$\Delta E=\frac{GMm}{r}\left(\frac{1}{2}-\frac{1}{3}\right)$
$\Delta E=\frac{GMm}{6r}$

To find the percentage increase with respect to the initial energy magnitude, we calculate:
$\text{Percentage Increase}=\frac{\Delta E}{\left|E_1\right|}\times 100$
$\text{Percentage Increase}=\frac{\frac{GMm}{6r}}{\frac{GMm}{2r}}\times 100$
$\text{Percentage Increase}=\frac{2}{6}\times 100=\frac{1}{3}\times 100\approx 33.3\%$

Therefore, the energy of the satellite must be increased by 33.3%.