Question 12.SP.12: A satellite is launched in a direction parallel to the surfa......
A satellite is launched in a direction parallel to the surface of the earth with a velocity of 18,820 mi/h from an altitude of 240 mi. Determine the velocity of the satellite as it reaches its maximum altitude of 2340 mi. Recall that the earth’s radius is 3960 mi.
STRATEGY: The satellite is acted on by a central force, so angular momentum is conserved. You can use the principle of conservation of angular momentum to determine the velocity of the satellite.
MODELING and ANALYSIS: Because the satellite is moving under a central force directed toward the center O of the earth, its angular momentum H_{O} is constant. From Eq. (12.14), you have
\mathbf{H}_{O}=r m\nu\,\mathrm{sin}\,\phi (12.14)
rmv sin ϕ = \mathbf{H}_{O} = constant
This equation shows that v is at a minimum at B, where both r and sin ϕ are maximum. Expressing the conservation of angular momentum between A and B, we have
r_{A}mv_{A}=r_{B}mv_{B}
Hence,
v_{B}=v_{A}{\frac{r_{A}}{r_{B}}}=(18,820\mathrm{~mi/h}){\frac{3960\mathrm{~mi}+240\mathrm{~mi}}{3960\mathrm{~mi}+2340\mathrm{~mi}}}
v_{B} = 12,550 mi/h ◂
REFLECT and THINK: Note that in order to increase velocity, you could choose to apply thrusters pushing the spacecraft closer to the earth. Because this is a central force, the spacecraft’s angular momentum remains constant. Therefore, its velocity v increases as the radial distance r decreases.
