Edits to “Statement”, “Solution”, “Answer”
en/2.6.40.md
+4 −4
| @@ -1,10 +1,10 @@ | |||
| ### Statement | |||
| − | $2.6.40.$ [Insert the problem statement] | ||
| + | $2.6.40.$ A space probe of mass $m$ moves around a planet of mass $M$ in an orbit with the greatest distance $r_a$ from the center of the planet (in the apocenter) and the smallest $r_p$ (in the pericenter). What is the minimum energy required for the probe to leave the planet? | ||
| ### Solution | |||
| − | From conservation of angular momentum, we have $m | ||
| + | From conservation of angular momentum, we have $mv_pr_p=mv_ar_a$, where $v_p$ and $v_a$ are the speeds at the pericenter and the apocenter, respectively. Together with the conservation of energy | ||
| \[\frac{1}{2}mv_p^2-\frac{GMm}{r_p}=\frac{1}{2}mv_a^2-\frac{GMm}{r_a},\] | |||
| we can solve for $v_p^2$ in terms of $r_p$ and $r_a$ and get | |||
| \[v_p^2=\frac{2GMr_a}{r_p(r_p+r_a)}.\] | |||
| Consequently, the energy of the space probe at the pericenter is | |||
| @@ -16,8 +16,8 @@Solution | |||
| \[\frac{1}{2}mv_p^2-\frac{GMm}{r_p}=-\frac{GMm}{r_p+r_a}.\] | |||
| − | Therefore, the space probe requires | ||
| + | Therefore, the space probe requires minimum energy of $GMm/(r_p+r_a)$ to escape from the planet (i.e. approaching infinity with very little kinetic energy left). | ||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | $\frac{GMm}{r_p+r_a}$ | ||
| @@ -1,10 +1,10 @@ | |||
| ### Statement | ### Statement | ||
| $2.6.40.$ [Insert the problem statement] | $2.6.40.$ A space probe of mass $m$ moves around a planet of mass $M$ in an orbit with the greatest distance $r_a$ from the center of the planet (in the apocenter) and the smallest $r_p$ (in the pericenter). What is the minimum energy required for the probe to leave the planet? | ||
| ### Solution | ### Solution | ||
| From conservation of angular momentum, we have $m |
From conservation of angular momentum, we have $mv_pr_p=mv_ar_a$, where $v_p$ and $v_a$ are the speeds at the pericenter and the apocenter, respectively. Together with the conservation of energy | ||
| \[\frac{1}{2}mv_p^2-\frac{GMm}{r_p}=\frac{1}{2}mv_a^2-\frac{GMm}{r_a},\] | \[\frac{1}{2}mv_p^2-\frac{GMm}{r_p}=\frac{1}{2}mv_a^2-\frac{GMm}{r_a},\] | ||
| we can solve for $v_p^2$ in terms of $r_p$ and $r_a$ and get | we can solve for $v_p^2$ in terms of $r_p$ and $r_a$ and get | ||
| \[v_p^2=\frac{2GMr_a}{r_p(r_p+r_a)}.\] | \[v_p^2=\frac{2GMr_a}{r_p(r_p+r_a)}.\] | ||
| Consequently, the energy of the space probe at the pericenter is | Consequently, the energy of the space probe at the pericenter is | ||
| @@ -16,8 +16,8 @@Solution | |||
| \[\frac{1}{2}mv_p^2-\frac{GMm}{r_p}=-\frac{GMm}{r_p+r_a}.\] | \[\frac{1}{2}mv_p^2-\frac{GMm}{r_p}=-\frac{GMm}{r_p+r_a}.\] | ||
| Therefore, the space probe requires |
Therefore, the space probe requires minimum energy of $GMm/(r_p+r_a)$ to escape from the planet (i.e. approaching infinity with very little kinetic energy left). | ||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | $\frac{GMm}{r_p+r_a}$ | ||