The mass of Jupiter is M=1.90×10^27 kg, and its radius is R=71500 km. A satellite is in a circular orbit around Jupiter at a radius of 3x Jupiter’s radius. A.) How does the satellite’s centripetal acceleration, a, compare to the acceleration due to gravity at the orbit, g? B.) What speed does the satellite have in its orbit? C.) What is the period of the orbit?
The Correct Answer and Explanation is:
We are given:
- Mass of Jupiter, M=1.90×1027 kgM = 1.90 \times 10^{27} \, \text{kg}
- Radius of Jupiter, R=71,500 km=7.15×107 mR = 71,500 \, \text{km} = 7.15 \times 10^7 \, \text{m}
- Orbit radius r=3R=3×7.15×107=2.145×108 mr = 3R = 3 \times 7.15 \times 10^7 = 2.145 \times 10^8 \, \text{m}
A) How does centripetal acceleration aa compare to gravitational acceleration gg?
Gravitational acceleration at the orbit: g=GMr2g = \frac{GM}{r^2}
Centripetal acceleration for circular motion: a=v2ra = \frac{v^2}{r}
For an object in orbit, the gravitational force provides the centripetal force: GMr2=v2r⇒a=g\frac{GM}{r^2} = \frac{v^2}{r} \Rightarrow a = g
✅ Answer A: The centripetal acceleration aa is equal to the gravitational acceleration gg at the orbit.
B) Speed of the satellite
v=GMr=(6.674×10−11)(1.90×1027)2.145×108≈5.915×108≈24310 m/sv = \sqrt{\frac{GM}{r}} = \sqrt{\frac{(6.674 \times 10^{-11})(1.90 \times 10^{27})}{2.145 \times 10^8}} \approx \sqrt{5.915 \times 10^8} \approx 24310 \, \text{m/s}
✅ Answer B: The speed is approximately 24,310 m/s\boxed{24,310 \, \text{m/s}}.
C) Orbital period TT
T=2πrv=2π(2.145×108)24310≈1.347×10924310≈55440 sT = \frac{2\pi r}{v} = \frac{2\pi (2.145 \times 10^8)}{24310} \approx \frac{1.347 \times 10^9}{24310} \approx 55440 \, \text{s}
Convert to hours: 554403600≈15.4 hours\frac{55440}{3600} \approx 15.4 \, \text{hours}
✅ Answer C: The period is approximately 15.4 hours\boxed{15.4 \, \text{hours}}.
Explanation
This problem concerns the motion of a satellite in a circular orbit around Jupiter, which is governed by Newtonian gravity and centripetal motion. In part A, we compare the satellite’s centripetal acceleration aa with the gravitational acceleration gg at its orbit. Since gravity is the force providing the necessary centripetal acceleration, these two quantities are always equal for a stable circular orbit. This is a foundational result in orbital mechanics, derived by equating gravitational force GMm/r2GMm/r^2 to the centripetal force mv2/rmv^2/r, leading to a=ga = g.
In part B, we calculate the orbital speed using the equation v=GM/rv = \sqrt{GM/r}. Substituting the known values gives a speed of about 24,310 m/s. This high speed is typical due to Jupiter’s large mass, which creates a strong gravitational pull even at a distance three times its radius.
Finally, in part C, we determine the orbital period using T=2πr/vT = 2\pi r / v, which represents the time to complete one full orbit. Plugging in the radius and velocity gives an orbital period of approximately 55,440 seconds, or about 15.4 hours.
This analysis demonstrates the interplay of gravitational and centripetal forces in planetary orbits. Despite the large radius, the satellite maintains high velocity and relatively short orbital period due to Jupiter’s immense mass. Such calculations are crucial in space mission planning, such as placing satellites or probes in orbit around large celestial bodies.
