Question

50POINTS! Find the period of revolution for the planet Mercury, whose average distance from the sun is 5.79e10. Mass of Sun is 1.99e30kg, mass of Mercury is 3.3e23kg

Answer 1

t=7.59(10^6)

Step-by-step explanation:

Answer 2

Answer:
`T=7.59(10^(6))s`

According to the Third Kepler’s Law of Planetary motion “The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

In other words, this law states a relation between the orbital period
`T` of a body (planet Mercury in this case) orbiting a greater body in space (the Sun) with the size
`r` of its orbit.

This Law is originally expressed as follows:

`T^(2)=(4\pi^(2))/(GM)r^(3)` (1)

Where;

`G` is the Gravitational Constant and its value is
`6.674x10^(-11)(m^(3))/(kgs^(2))`

`M=1.99×10^(30)kg` is the mass of the Sun

`r=5.79×10^(10)m is the <strong>distance from Mercury to the Sun</strong> (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit). </p><p> </p><p>If we want to find the period, we have to express equation<strong> (1)</strong> as written below and substitute all the values: </p><p> </p><p>[tex]T=\sqrt{(4\pi^(2))/(GM)r^(3)}` (2)

`T=2\pi\sqrt{(r^(3))/(GM)` (3)

`T=2\pi\sqrt{((5.79(10^(10)m))^(3))/((6.674x10^(-11)(m^(3))/(kgs^(2)))(1.99×10^(30)kg))` (4)

Solving and taking into account that
`1N=1kg(m)/(s^(2))`:

`T=2\pi\sqrt{1.46(10^(11))s^(2)}` (5)

Finally:

`T=7.59(10^(6))s`>>>>This is the period of revolution for the planet Mercury