Question

NASA has asked your team of rocket scientists about the feasibility of a new satellite launcher that will save rocket fuel. NASA's idea is basically an electric slingshot that consists of 4 electrodes arranged in a horizontal square 5 m on a side. The satellite is placed 15 m directly under the center of the square. A power supply will provide each of the four electrodes with the same charge and the satellite with an opposite charge 4 times larger. When the satellite is released from rest, it moves up and passes through the center of the square. At the instant it reaches the square's center, the power supply is turned off and the electrodes are grounded, giving them a zero electric charge. To test this idea, you decide to use energy considerations to calculate the electrode charge necessary to get a 100 kg satellite to an orbit height of 300 km. In your physics text you find the mass of the Earth to be 6.0 × 1024 kg and the radius of the Earth to be about 6,400 km. As you begin the calculations, a teammate asks, “Hey, wait. Do they want us to just launch it up that high? Or do they want us to launch it so it gets that high with the right velocity to go into a circular orbit?” Make sure to address your teammate’s question in your Explain step.

Answer 1

Initial potential energy when the satellite( charge 4 q ) is at a distance of 15 m below from the center of squre formed by electrodes (charge q ) is given by $U_i=\frac{1}{4\pi\epsilon_0}\frac{4\times4q^2}{\sqrt{(15)^2+(5/\sqrt2)^2}}$

Final potential energy when the satellite is at the center of square formed by electrodes is given by

$U_f=\frac{1}{4\pi\epsilon_0}\,\frac{4\times4q^2}{(5/\sqrt2)}$

change in potential energy = $U_f -U_i =\frac{16q^2}{4\pi\epsilon_0}\left ( \frac{\sqrt2}{5}-\frac{1}{\sqrt{(15)^2+(5/\sqrt2)^2}}\right )=q^2\times31.35 \times10^9\,J$

This energy is equal to the energy needed to raise a satellite of mass 100 kg to a height 300 km above the surface of earth.

$=\frac{GMm}{R}-\frac{GMm}{R+h}$

$=(6.67\times10^{-11}\times6.0\times10^{24}\times100)\left(\frac{1}{6400\times1000}-\frac{1}{6700\times1000} \right )=$

$=0.02799\times10^{10}\,J$

$q^2\times31.35\times10^{9}=0.02799\times10^{10}$

$q=0.0945\,C$

This arrangement sends the satellite to a height 300 km, but does not put it in to orbit.

To put the satellite in to orbit some other arrangement is necessary.