Question

An artificial satellite is in a circular orbit around a planet of radius r= 2.05 x103 km at a distance d 310.0 km from the planet's surface. The period of revolution of the satellite around the planet is T 1.15 hours. What is the average density of the planet?

Answer 1

`\rho = 12580.7 kg/m^3`

Step-by-step explanation:

As we know that the satellite revolves around the planet then the centripetal force for the satellite is due to gravitational attraction force of the planet

So here we will have

`F = (GMm)/((r + h)^2)`

here we have

`F =\frac {mv^2}{(r+ h)}`

`(mv^2)/(r + h) = (GMm)/((r + h)^2)`

here we have

`v = \sqrt{(GM)/((r + h))}`

now we can find time period as

`T = (2\pi (r + h))/(v)`

`T = \frac{2\pi (2.05 * 10^6 + 310 * 10^3)}{\sqrt{(GM)/((r + h))}}`

`1.15 * 3600 = \frac{2\pi (2.05 * 10^6 + 310 * 10^3)}{\sqrt{((6.67 * 10^(-11))(M))/((2.05 * 10^6 + 310 * 10^3))}}`

`M = 4.54 * 10^(23) kg`

Now the density is given as

`\rho = (M)/((4)/(3)\pi r^3)`

`\rho = \frac{4.54 * 10^(23)}{\frac{4}[3}\pi(2.05 * 10^6)^3}`

`\rho = 12580.7 kg/m^3`