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2 years ago

From the average distance of one of jupiter's moons to jupiter and its orbital period around jupiter, we can determine:

2 answers:

5 0

The mass of the Jupiter can be easily calculated using the average distance of the Jupiter's moon and its orbital period around the planet.

Explanation:

The Kepler's laws of Planetary motion explain about the several parameters of the planet and the moon system. The three laws of planetary motion explain about the nature of the orbit, orbital velocity and also about the time period of the moon around the planet.

The Kepler's third Law of planetary motion establishes the relation between the distance between the planet and the moon or the orbital distance and the time taken by the moon in completing one rotation around the planet.

The mathematical expression for the Kepler's third law is given as:

$\boxed{T^2=\dfrac{4\pi^2}{GM}R^3}$

Here, $T$ is the time period of the moon, $G$ is the gravitational constant, $M$ is the mass of the planet and $R$ is the orbital distance of the moon from the planet.

In case of the Jupiter and its moon, if the orbital distance of the moon and the time period of rotation of the moon around the planet is known, then the mass of the planet Jupiter can be easily determined by means of the equation of the third law of Kepler.

Thus, The mass of the Jupiter can be easily calculated using the average distance of the Jupiter's moon and its orbital period around the planet.

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Answer Details:

Grade: High School

Subject: Physics

Chapter: Gravitation

Keywords:

Average distance, Jupiter, moon, Kepler's law, planetary motion, third law, orbital distance, mass, gravitational constant, time period, rotation.

den301095 [7]2 years ago

4 0

Answer: Jupiter's mass

Explanation:

From Kepler's third law:

$T^2=\frac{4\pi^2}{GM}a^3$

where T is the orbital period of a satellite, a is the average distance of the satellite from the Planet, M is the mass of the planet, G is the gravitational constant.

If the average distance of one of Jupiter's moons to Jupiter and its orbital period around Jupiter is given then mass of the Jupiter can be found:

$\Rightarrow M_J=\frac{4\pi^2}{GT_m^2}a_m^3$

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We said that the speed at the top of the trajectory (part a) is

$v_1 = 18.9 m/s$

while the speed at half of the maximum height (part b) is

$v_2 = 22.2 m/s$

So the difference is

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And so, in percentage,

$\frac{\Delta v}{v_1} \cdot 100 = \frac{3.3}{18.9}\cdot 100=17.4\%$

So, the stone in part (b) is moving 17.4% faster than in part (a).

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