Find the mass of Jupiter based on data for the orbit of one of its moons, and compare your result with its actual mass.

Short Answer

Expert verified

Jupiter’s mass is 1.90×1027 kg.

The calculated value agrees with the actual value for the mass of Jupiter.

Step by step solution

01

Definition of Force

A force is a factor that can impact an object's motion. A force can cause a mass item to accelerate (e.g., from a standstill).

02

Calculation of the centripetal force on the Jupiter’s moon Europa due to revolution

The gravitational force between the two objects is given by,

Fg=Gm1m2r2

Where G is the universal gravitation constant, r is the distance between the two objects, and m1,m2are the masses of two bodies, respectively.

In angular motion, the centripetal acceleration is given,

Fc=mac=mω2r

As the moon revolves around Jupiter, so there is the only force that is the gravitational force which provides centripetal force to the moon for revolution. Therefore, the centripetal force on the moon is equal to the gravitational force between Jupiter and its moon.

Mathematically, it can be written as,

Fg=FcGmearthmsunr2=mearthω2rmsun=ω2r3G

03

Calculate the mass of Jupiter

In 3.55 days, the Europa completes one revolution of Jupiter. So, the angular velocity of the Europa is,

ω=1 rev3.55 daysω=1 rev3.55 d×2π rad1 rev×1 d24 hr×1 hr3600 sω=2.05×10-5rads

The distance of Jupiter from one of its moons, Europa, in Table 6.2 is r=6.71×108m.

So, the mass of the Jupiter,

msun=ω2r3G=2.05×10-52×6.71×10836.673×10-11msun=1.90×1027 kg

Hence, the mass of Jupiter based on the data from one of its moons, Europa orbit is 1.90×1027 kg, which is very close to Jupiter’s actual mass that is 1.898×1027 kg. So, the calculated value agrees with the actual value for the mass of Jupiter.

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Most popular questions from this chapter

Action at a distance, such as is the case for gravity, was once thought to be illogical and therefore untrue. What is the ultimate determinant of the truth in physics, and why was this action ultimately accepted?

(a) Calculate the magnitude of the acceleration due to gravity on the surface of Earth due to the Moon. (b) Calculate the magnitude of the acceleration due to gravity at Earth due to the Sun. (c) Take the ratio of the Moon’s acceleration to the Sun’s and comment on why the tides are predominantly due to the Moon in spite of this number.

A large centrifuge, like the one shown in Figure (a), is used to expose aspiring astronauts to accelerations similar to those experienced in rocket launches and atmospheric re-entries. (a) At what angular velocity is the centripetal acceleration \(10g\) if the rider is \(15.0{\rm{ m}}\) from the centre of rotation? (b) The rider’s cage hangs on a pivot at the end of the arm, allowing it to swing outward during rotation as shown in Figure (b). At what angle \(\theta \) below the horizontal will the cage hang when the centripetal acceleration is \(10g\)? (Hint: The arm supplies centripetal force and supports the weight of the cage. Draw a free body diagram of the forces to see what the angle \(\theta \) should be.)



*Figure (a) NASA centrifuge used to subject trainees to accelerations similar to those experienced in rocket launches and re-entries. (Credit: NASA) (b) Rider in cage showing how the cage pivots outward during rotation. This allows the total force exerted on the rider by the cage to be along its axis at all times.

Suppose a mass is moving in a circular path on a frictionless table as shown in figure. In the Earth’s frame of reference, there is no centrifugal force pulling the mass away from the centre of rotation, yet there is a very real force stretching the string attaching the mass to the nail. Using concepts related to centripetal force and Newton’s third law, explain what force stretches the string, identifying its physical origin.

Part of riding a bicycle involves leaning at the correct angle when making a turn, as seen in Figure. To be stable, the force exerted by the ground must be on a line going through the center of gravity. The force on the bicycle wheel can be resolved into two perpendicular components—friction parallel to the road (this must supply the centripetal force), and the vertical normal force (which must equal the system’s weight).

(a) Show that\(\theta \)(as defined in the figure) is related to the speed v and radius of curvature r of the turn in the same way as for an ideally banked roadway—that is,\(\theta = {\tan ^{ - 1}}\,{v^2}/rg\)

(b) Calculate \(\theta \) for a \(12.0{\rm{ m}}/{\rm{s}}\) turn of radius \(30.0{\rm{ m}}\) (as in a race).

Figure 6.36 A bicyclist negotiating a turn on level ground must lean at the correct angle—the ability to do this becomes instinctive. The force of the ground on the wheel needs to be on a line through the center of gravity. The net external force on the system is the centripetal force. The vertical component of the force on the wheel cancels the weight of the system while its horizontal component must supply the centripetal force. This process produces a relationship among the angle \(\theta \), the speed \(v\), and the radius of curvature \(r\) of the turn similar to that for the ideal banking of roadways.

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