Asatellite in orbit around the earth has a period of one hour. an identical satellite is placed in an orbit having a radius that is nine times larger than that of the first satellite. what is the period of the second satellite?

27h

Explanation:

We can answer this problem with Kepler's third law:

where is the period and is the semi-major axis of an ellipse, but because the satellite's orbit is around the earth it must be a circular orbit, so that becomes the radius of a circle.

what this tells us is that the relationship between the first period with radius must be equal to the second period when the radius is now 9 times the original (the second radius is because the orbit is nine times larger):

We will have the following:

is the original period: 1 hour

is the period of the second satellite.

and

thus:

Clearing for

The period of the second satellite is 27 hours

powerful computers now, not then

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j