What is the electric field 4.0 m from the center of the terminal of a Van de Graaff with a 7.10 mC charge, noting that the field
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<h2>Answer: 2T</h2>
According to the Third Kepler’s Law of Planetary motion <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 (in the case of planet X and assuming we have a circular orbit):
(1)
Where:
is the Gravitational Constant
is the mass of planet X
is the radius of the orbit of the satellite around planet X
If we want to find the period, we have to express equation (1) as written below and substitute all the values:
(2)
Now, we are asked to find the period when tha mass of the planet is . In order to do this, we have to rewrite equation (2) with this new value:
(3)
Solving:
(4)
On the other hand, if we multiply both sides of equation (2) by 2, we have:
(5)
As we can see, (5) is equal to (4). This means the orbital period is twice the orignal period.
Hence, the answer is:
If Planet X had <u>one-fourth </u>as much mass, the <u>orbital period</u> of this satellite in an orbit of the same radius would be <u>2T.</u>