Answer:
[tex]r_{2} =\frac{80,000 \;km}{4} =20,000 \;km[/tex]
Explanation:
To find the new orbital radius where the satellite's velocity is doubled, we'll need to use the relationship between orbital radius and velocity for circular orbits, which is derived from Kepler's Third Law and Newton's Law of Gravitation.
Original Orbit Details:
Orbital Velocity Equation:
The orbital velocity ([tex]v[/tex]) in a circular orbit is given by:
[tex]v=\sqrt\frac{GM}{r}[/tex]
where:
[tex]G[/tex] is the gravitational constant [tex](6.67430 \; \times 10^{-11} \; m^{3} kg^{-1} s^{-2} )[/tex]
[tex]M[/tex] is the mass of the Earth [tex](5.972 \times 10^{24} \,kg)[/tex]
[tex]r[/tex] is the orbital radius
Initial Velocity Calculation:
[tex]v_{1} =\sqrt\frac{GM}{r_{1} }[/tex]
New Velocity:
The new velocity [tex]v_{2}[/tex] is double the initial velocity:
[tex]v_{2} = 2 v_{1}[/tex]
New Orbital Radius Calculation:
Using the orbital velocity formula again for the new orbit:
[tex]v_{2} =\sqrt\frac{GM}{r_{2} }[/tex]
Since [tex]v_{2} =2v_{1}[/tex]:
[tex]2v_{1}= \sqrt\frac{GM}{r_{2} }[/tex]
Substituting [tex]v_{1}[/tex] from the initial velocity equation:
[tex]2\sqrt\frac{GM}{r_{1} } =\sqrt\frac{GM}{r_{2} }[/tex]
Solving for the New Radius:
Square both sides to remove the square roots:
[tex]4\sqrt\frac{GM}{r_{1} }=\sqrt\frac{GM}{r_{2} }[/tex]
Cancel out from both sides:
[tex]4\frac{r}{1} =\frac{1}{r_{2} }[/tex]
Therefore:
[tex]r_{2} =\frac{r_{1} }{4}[/tex]
Final Calculation:
Given [tex]r_{1}[/tex] = 80,000 km:
[tex]r_{2} =\frac{80,000\, km}{4} = 20,000 \, km[/tex]