(A) The expression for the orbital speed of satellite is [tex]v =\dfrac{\sqrt{2 \times G \times mp}}{a}[/tex].
(B) The energy of satellite is [tex]E = \dfrac{G \times ms \times mp}{a^{2}}[/tex].
Given data:
The mass of satellite is, ms.
The mass of planet is, mp.
The radius of planet is, a.
(A)
We need to find the orbital speed of satellite. So clearly we known that the while going round the planet, a satellite is experienced with centripetal force, balanced by the gravitational force.
So,
Fc = Fg
Here, Fc is the centripetal force and Fg is the gravitational force.
[tex]\dfrac{ms \times v^{2}}{2}=\dfrac{G \times ms \times mp}{a^{2}}[/tex]
G is the universal gravitational constant and v is the orbital speed of satellite.
Solving as,
[tex]\dfrac{ v^{2}}{2}=\dfrac{G \times mp}{a^{2}}\\\\\\v = \sqrt{\dfrac{2 \times G \times mp}{a^{2}}}\\\\\\v = \sqrt{\dfrac{2 \times G \times mp}{a^{2}}}\\\\\\v =\dfrac{\sqrt{2 \times G \times mp}}{a}[/tex]
Thus, we can conclude that the expression for the orbital speed of satellite is [tex]v =\dfrac{\sqrt{2 \times G \times mp}}{a}[/tex].
(B)
The energy of the satellite is nothing but the kinetic energy of satellite. Then the required energy of satellite is,
[tex]E = \dfrac{1}{2} \times ms \times v^{2}\\\\E = \dfrac{1}{2} \times ms \times \dfrac{2G \times mp}{a^{2}}\\\\\\E = \dfrac{G \times ms \times mp}{a^{2}}[/tex]
Thus, we can conclude that the energy of satellite is [tex]E = \dfrac{G \times ms \times mp}{a^{2}}[/tex].
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