Answer:
a) Greater than that for an Earth satellite.
Explanation:
The orbital speed of a spacecraft in a low orbit around a celestial body depends on the gravitational acceleration at that location. The formula for orbital speed ([tex]\sf v[/tex]) is given by:
[tex]\boxed{\boxed{\sf v = \sqrt{\dfrac{G \cdot M}{R}}}} [/tex]
where:
[tex]\sf v[/tex] is the orbital speed,
[tex]\sf G[/tex] is the gravitational constant ([tex]\sf 6.67430 \times 10^{-11} \ \textsf{m}^3 \ \textsf{kg}^{-1} \ \textsf{s}^{-2}[/tex]),
[tex]\sf M[/tex] is the mass of the celestial body,
[tex]\sf R[/tex] is the distance from the center of the celestial body to the spacecraft.
In this case, we know that the free-fall acceleration ([tex]\sf g[/tex]) near Jupiter's surface is 2.5 times that on Earth, and the radius of Jupiter ([tex]\sf R_{\textsf{Jupiter}}[/tex]) is 11 times that of Earth ([tex]\sf R_{\textsf{Earth}}[/tex]).
[tex]\sf g_{\textsf{Jupiter}} = 2.5 \cdot g_{\textsf{Earth}} [/tex]
[tex]\sf R_{\textsf{Jupiter}} = 11 \cdot R_{\textsf{Earth}} [/tex]
Now, the gravitational acceleration is related to the orbital speed by the formula:
[tex]\sf g = \dfrac{v^2}{R} [/tex]
We can express [tex]\sf v[/tex] in terms of [tex]\sf g[/tex] and [tex]\sf R[/tex]:
[tex]\sf v = \sqrt{g \cdot R} [/tex]
For Jupiter:
[tex]\sf g_{\textsf{Jupiter}} = \dfrac{v_{\textsf{Jupiter}}^2}{R_{\textsf{Jupiter}}} [/tex]
Substitute the given relationships:
[tex]\sf 2.5 \cdot g_{\textsf{Earth}} = \dfrac{v_{\textsf{Jupiter}}^2}{11 \cdot R_{\textsf{Earth}}} [/tex]
Now, let's compare the orbital speed of the spacecraft around Jupiter ([tex]\sf v_{\textsf{Jupiter}}[/tex]) with the orbital speed of an Earth satellite ([tex]\sf v_{\textsf{Earth}}[/tex]):
[tex]\sf v_{\textsf{Jupiter}} = \sqrt{2.5 \cdot g_{\textsf{Earth}} \cdot 11 \cdot R_{\textsf{Earth}}} [/tex]
[tex]\sf v_{\textsf{Jupiter}} = \sqrt{2.5 \cdot v_{\textsf{Earth}}^2} [/tex]
[tex]\sf v_{\textsf{Jupiter}} = \sqrt{2.5} \cdot v_{\textsf{Earth}} [/tex]
Since [tex]\sf \sqrt{2.5}[/tex] is greater than 1, the orbital speed around Jupiter ([tex]\sf v_{\textsf{Jupiter}}[/tex]) is greater than the orbital speed of an Earth satellite ([tex]\sf v_{\textsf{Earth}}[/tex]).
Therefore, the correct answer is:
a) Greater than that for an Earth satellite.
