Answer with explanation: Provided the satellitle orbiting the earth ( Planet ), we have to fill the missing information in the blanks:
Acceleration of the satellite:
[tex]\begin{gathered} a=-g \\ g=\frac{M}{r^2}G \\ G=6.674\cdot10^{-11}\cdot m^3\cdot kg^{-1}\cdot s^{-2} \\ r=(5.81\times10^7)m \\ M=(5.97\times10^{24})\operatorname{kg} \\ \therefore\Rightarrow \\ g=\frac{(5.97\times10^{24})\operatorname{kg}}{(5.81\times10^7m)^2}\cdot(6.674\cdot10^{-11}\cdot m^3\cdot kg^{-1}\cdot s^{-2}) \\ g=-6.8578\cdot10^6ms^{-2} \\ a=-g \\ \therefore\Rightarrow \\ a=-(-6.8578\cdot10^6ms^{-2}_{})= \\ a=6.8578\cdot10^6ms^{-2} \end{gathered}[/tex]Value of G-Constant:
[tex]G=6.674\cdot10^{-11}\cdot m^3\cdot kg^{-1}\cdot s^{-2}[/tex]Mass of Earth:
[tex]M=(5.97\times10^{24})\operatorname{kg}[/tex]Radius or distance between the earth and the satellite:
[tex]r=(5.81\times10^7)m[/tex]The velocity of the satellite:
[tex]\begin{gathered} a=\frac{v^2}{r} \\ \therefore\Rightarrow \\ v=\sqrt[]{ra} \\ v=\sqrt[]{(5.81\times10^7)m\cdot(6.8578\cdot10^6ms^{-2}}) \\ v=7.9520\times10^7 \end{gathered}[/tex]Time period T:
[tex]\begin{gathered} T=\frac{2\pi r}{v} \\ \pi=3.141 \\ r(5.81\times10^7)m \\ v=7.9520\times10^7\therefore\Rightarrow \\ T=\frac{2\cdot\pi\cdot(5.81\times10^7)m}{(7.9520\times10^7)}=4.5898s \\ \therefore\Rightarrow \\ T=4.5898s\frac{1hr}{3600s}=0.0012749hr \\ T=0.0012749hr \end{gathered}[/tex]Following is the screenshot of the answer, as per your own request.
