QuestionDecember 13, 2025

The radius of Jupiter is 11 times that of earth, and the free-fall acceleration near its surface is 2.5 times that on earth If we someday put a spacecraft in low Jupiter orbit, its orbital speed will be greater than that for an earth satellite. less than that for an earth satellite the same as that for an earth satellite

The radius of Jupiter is 11 times that of earth, and the free-fall acceleration near its surface is 2.5 times that on earth If we someday put a spacecraft in low Jupiter orbit, its orbital speed will be greater than that for an earth satellite. less than that for an earth satellite the same as that for an earth satellite
The radius of Jupiter is 11 times that of earth, and the free-fall acceleration near its surface is 2.5 times that on earth If we someday put a spacecraft in low Jupiter orbit, its orbital speed will be greater than that for an earth satellite. less than that for an earth satellite the same as that for an earth satellite

Solution
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Answer

Greater than that for an earth satellite. Explanation 1. Express Orbital Speed Formula The orbital speed v for a low (near-surface) circular orbit around a planet of mass M and radius R is given by v = \sqrt{\frac{GM}{R}}, where G is the gravitational constant. 2. Relate Mass and Radius to Surface Gravity The surface gravity g is g = \frac{GM}{R^2}. Rearranging gives GM = gR^2. Substitute this into the orbital speed formula: v = \sqrt{\frac{gR^2}{R}} = \sqrt{gR}. 3. Compare Jupiter and Earth Orbital Speeds Let g_J and R_J be Jupiter's surface gravity and radius, g_E and R_E Earth's. Then, \[ \frac{v_J}{v_E} = \frac{\sqrt{g_J R_J}}{\sqrt{g_E R_E}} = \sqrt{\frac{g_J}{g_E} \cdot \frac{R_J}{R_E}} \] Given g_J = 2.5g_E and R_J = 11R_E, \[ \frac{v_J}{v_E} = \sqrt{2.5 \times 11} = \sqrt{27.5} \approx 5.24 \] So, the orbital speed near Jupiter is much greater than that near Earth.

Explanation

1. Express Orbital Speed Formula<br /> The orbital speed $v$ for a low (near-surface) circular orbit around a planet of mass $M$ and radius $R$ is given by $v = \sqrt{\frac{GM}{R}}$, where $G$ is the gravitational constant.<br />2. Relate Mass and Radius to Surface Gravity<br /> The surface gravity $g$ is $g = \frac{GM}{R^2}$. Rearranging gives $GM = gR^2$. Substitute this into the orbital speed formula: $v = \sqrt{\frac{gR^2}{R}} = \sqrt{gR}$.<br />3. Compare Jupiter and Earth Orbital Speeds<br /> Let $g_J$ and $R_J$ be Jupiter's surface gravity and radius, $g_E$ and $R_E$ Earth's. Then,<br />\[<br />\frac{v_J}{v_E} = \frac{\sqrt{g_J R_J}}{\sqrt{g_E R_E}} = \sqrt{\frac{g_J}{g_E} \cdot \frac{R_J}{R_E}}<br />\]<br />Given $g_J = 2.5g_E$ and $R_J = 11R_E$,<br />\[<br />\frac{v_J}{v_E} = \sqrt{2.5 \times 11} = \sqrt{27.5} \approx 5.24<br />\]<br />So, the orbital speed near Jupiter is much greater than that near Earth.
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