Question

How do I solve this problem? Hint:To find the satellite's acceleration, speed and period, use the following equations:a = G M / r2v = √(a r)T = 2πr/vBe sure to use the distance between the satellite and the earth for the radius! a=gravitational acceleration [m/s2]G=gravitational constant [always 6.67 x 10-11 m3/kg s2]M=larger mass [kg]r=distance between masses or orbital radius [m]v=orbital velocity or speed [m/s]T=period / time required to complete one revolution [s]The period can be converted to hours by using the conversion factor: 1 hour = 3600s.

Answer 1

Answer with explanation: Provided the satellitle orbiting the earth ( Planet ), we have to fill the missing information in the blanks:

Acceleration of the satellite:

Value of G-Constant:

`G=6.674\cdot10^(-11)\cdot m^3\cdot kg^(-1)\cdot s^(-2)`

Mass of Earth:

Radius or distance between the earth and the satellite:

`r=(5.81*10^7)m`

The velocity of the satellite:

`\begin{gathered} a=(v^2)/(r) \\ \therefore\Rightarrow \\ v=\sqrt[]{ra} \\ v=\sqrt[]{(5.81*10^7)m\cdot(6.8578\cdot10^6ms^(-2)}) \\ v=7.9520*10^7 \end{gathered}`

Time period T:

`\begin{gathered} T=(2\pi r)/(v) \\ \pi=3.141 \\ r(5.81*10^7)m \\ v=7.9520*10^7\therefore\Rightarrow \\ T=(2\cdot\pi\cdot(5.81*10^7)m)/((7.9520*10^7))=4.5898s \\ \therefore\Rightarrow \\ T=4.5898s(1hr)/(3600s)=0.0012749hr \\ T=0.0012749hr \end{gathered}`

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