According to the <u>Third Kepler’s Law of Planetary motion</u> “<em>The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>
In other words, this law states a relation between the orbital period 
of a body (moon, planet, satellite) orbiting a greater body in space with the size 
of its orbit.
This Law is originally expressed as follows:
<h2>
(1)
</h2>
Where;
is the Gravitational Constant and its value is 
is the mass of Jupiter
is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)
If we want to find the period, we have to express equation (1) as written below and substitute all the values:
<h2>
(2)
</h2>
Then:
<h2>
(3)
</h2>
Which is the same as:
<h2>
</h2>
Therefore, the answer is:
The orbital period of Io is 42.482 h
Answer: a) Mr = 2.4×10^-4kg/s
V = 34.42m/a
b) E = 173J
Ø = 2693.1J
c) Er = 0.64J/s
Explanation: Please find the attached file for the solution
Answer:
The horizontal component of the velocity is 21.9 m/s.
Explanation:
Please see the attached figure for a better understanding of the problem.
Notice that the vector v and its x and y-components (vx and vy) form a right triangle. Then, we can use trigonometry to find the magnitude of vx, the horizontal component of the velocity.
To find vx, let´s use the following trigonometric rule of right triangles:
cos α = adjacent / hypotenuse
cos 5.7° = vx / 22 m/s
22 m/s · cos 5.7° = vx
vx = 21.9 m/s
The horizontal component of the velocity is 21.9 m/s.
6 m/s because the position of the object was increasing 6 m every second
Answer:
0 J
Explanation:
From the diagram below; we would notice that the Force (F) = Tension (T)
Also the angle θ adjacent to the perpendicular line = 90 °
The Workdone W = F. d
W = Fd cos θ
W = Fd cos 90°
W = Fd (0)
W = 0 J
Hence the force is perpendicular to the direction of displacement and the net work done in a circular motion in one complete revolution is = 0
