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Tete edited Aug 27, 2026, 1:50 AM

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		+	### Statement
		+
		+	$2.6.40.$ [Insert the problem statement]
		+
		+	### Solution
		+
		+	From conservation of angular momentum, we have $m_v_pr_p=mv_ar_a$, 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
		+
		+	\[\frac{1}{2}mv_p^2-\frac{GMm}{r_p}=-\frac{GMm}{r_p+r_a}.\]
		+
		+	Therefore, the space probe requires a 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]

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			### Statement

			$2.6.40.$ [Insert the problem statement]

			### Solution

			From conservation of angular momentum, we have $m_v_pr_p=mv_ar_a$, 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

			\[\frac{1}{2}mv_p^2-\frac{GMm}{r_p}=-\frac{GMm}{r_p+r_a}.\]

			Therefore, the space probe requires a 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]