Solve the problem.
-Find the values of $v _ { 0 }$ in the equation $e = \frac { r _ { 0 } v _ { 0 } ^ { 2 } } { G M } - 1$ that make the orbit described by $r = \frac { ( 1 + e ) r _ { 0 } } { 1 + e \cos \theta }$ parabolic.
A) $v _ { 0 } = 2 G M$
B) $v _ { 0 } = \frac { 2 G M } { r _ { 0 } }$
C) $v _ { 0 } = \sqrt { \frac { 2 G M } { r _ { 0 } } }$
D) $\mathrm { v } _ { 0 } = \sqrt { 2 \mathrm { GMr } 0 }$

Question

Quizplus · Accepted Answer

For a parabolic orbit, the eccentricity is equal to 1. So, we set $e = 1$ and solve for $v_0$:
\begin{align*}
1 &= \frac{r_0v_0^2}{GM} - 1\
2 &= \frac{r_0v_0^2}{GM}\
v_0 &= \sqrt{\frac{2GM}{r_0}}
\end{align*}
Thus, the answer is (C).

Solve the Problem v0v _ { 0 }v0​ In the Equation e=r0v02GM−1e = \frac { r _ { 0 } v _ { 0 } ^ { 2 } } { G M } - 1e=GMr0​v02​​−1

Solve the Problem $v _ { 0 }$ In the Equation $e = \frac { r _ { 0 } v _ { 0 } ^ { 2 } } { G M } - 1$