Answer
Mercury's distance from the sun at the closest point is $4.60\times 10^{10}~m$
Work Step by Step
We can use conservation of angular momentum to find Mercury's speed at its closest point. Let $L_c$ be the angular momentum at the closest point. Let $L_f$ be the angular momentum at the farthest point. Let $M_m$ be Mercury's mass.
$L_c = L_f$
$M_m~v_c~r_c ~sin(90^{\circ})= M_m~v_f~r_f~sin(90^{\circ})$
$r_c = \frac{v_f~r_f}{v_c}$
$r_c = \frac{(38.8~km/s)(6.99\times 10^{10}~m)}{59.0~km/s}$
$r_c = 4.60\times 10^{10}~m$
Mercury's distance from the sun at the closest point is $4.60\times 10^{10}~m$.
