Respuesta :
Answer:
[tex]\rho = 12580.7 kg/m^3[/tex]
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
[tex]F = \frac{GMm}{(r + h)^2}[/tex]
here we have
[tex]F =\frac {mv^2}{(r+ h)}[/tex]
[tex]\frac{mv^2}{r + h} = \frac{GMm}{(r + h)^2}[/tex]
here we have
[tex]v = \sqrt{\frac{GM}{(r + h)}}[/tex]
now we can find time period as
[tex]T = \frac{2\pi (r + h)}{v}[/tex]
[tex]T = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{GM}{(r + h)}}}[/tex]
[tex]1.15 \times 3600 = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{(6.67 \times 10^{-11})(M)}{(2.05 \times 10^6 + 310 \times 10^3)}}}[/tex]
[tex]M = 4.54 \times 10^{23} kg[/tex]
Now the density is given as
[tex]\rho = \frac{M}{\frac{4}{3}\pi r^3}[/tex]
[tex]\rho = \frac{4.54 \times 10^{23}}{\frac{4}[3}\pi(2.05 \times 10^6)^3}[/tex]
[tex]\rho = 12580.7 kg/m^3[/tex]
