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vlabodo [156]
3 years ago
13

Two planets A and B, where B has twice the mass of A, orbit the Sun in elliptical orbits. The semi-major axis of the elliptical

orbit of planet B is two times larger than the semi major axis of the elliptical orbit of planet A. What is the ratio of the orbital period of planet B to that of planet A?
Physics
2 answers:
erastova [34]3 years ago
8 0

Answer:

T_b / T_a = sqrt ( 8 )    

Explanation:

Given:

- The mass of planet A = m _a

- The mass of planet B = 2*m _a

- The semi-major axis of plant A = a

- The semi-major axis of plant B = 2*a

Find:

- What is the ratio of the orbital period of planet B to that of planet A?

Solution:

- Kepler's Third Law for planetary motion of a satellite in an elliptical orbit is gives us the relation of time period as follows:

                            T = 2*pi*sqrt ( a^3 / G*M_s )

Where,

a : Semi major axis of the period.

M_s : It is the mass of the sun.

G: Gravitational constant

- So for planet A, we can develop an expression for time period:

                            T_a = 2*pi*sqrt ( a^3 / G*M_s )

- And for planet B, we can develop an expression for time period:

                            T_b = 2*pi*sqrt ( (2a)^3 / G*M_s )

- Now take the ratio of T_b to T_a:

     T_b / T_a = 2*pi*sqrt ( (2a)^3 / G*M_s ) / 2*pi*sqrt ( a^3 / G*M_s )

Simplify,

                T_b / T_a = sqrt ( (2a)^3 / a^3 ) = sqrt ( 8a^3 / a^3 )

                                     T_b / T_a = sqrt ( 8 )              

lozanna [386]3 years ago
6 0

Answer:

2.83

Explanation:

Kepler's discovered that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, that is called Kepler's third law of planet motion and can be expressed as:

T=\frac{2\pi a^{\frac{3}{2}}}{\sqrt{GM}} (1)

with T the orbital period, M the mass of the sun, G the Cavendish constant and a the semi major axis of the elliptical orbit of the planet. By (1) we can see that orbital period is independent of the mass of the planet and depends of the semi major axis, rearranging (1):

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

\frac{T^{2}}{a^{3}}=(\frac{2\pi }{\sqrt{GM}})^2 (2)

Because in the right side of the equation (2) we have only constant quantities, that implies the ratio \frac{T^{2}}{a^{3}} is constant for all the planets orbiting the same sun, so we can said that:

\frac{T_{A}^{2}}{a_{A}^{3}}=\frac{T_{B}^{2}}{a_{B}^{3}}

\frac{T_{B}^{2}}{T_{A}^{2}}=\frac{a_{B}^{3}}{a_{A}^{3}}

\frac{T_{B}}{T_{A}}=\sqrt{\frac{a_{B}^{3}}{a_{A}^{3}}}=\sqrt{\frac{(2a_{A})^{3}}{a_{A}^{3}}}

\frac{T_{B}}{T_{A}}=\sqrt{\frac{2^3}{1}}=2.83

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Given the box resting on the inclined plane above has a mass of 20kg and the The incline sits at a 30 degree angle

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