Respuesta :
Answer:
Part a)
[tex]\frac{F_1}{F_2} = 10.125[/tex]
Part b)
[tex]\frac{v_1}{v_2} = \sqrt{\frac{4.5r}{r}} = 2.12[/tex]
Part c)
[tex]\frac{T_1}{T_2} = 9.54[/tex]
Explanation:
Part a)
As we know that the gravitational force is given as
[tex]F = \frac{GMm}{r^2}[/tex]
so we will have to find the ratio of force on two planets due to star
so here we have
[tex]\frac{F_1}{F_2} = \frac{m_1r_2^2}{m_2r_1^2}[/tex]
[tex]\frac{F_1}{F_2} = \frac{m (4.5r)^2}{(2m) r}[/tex]
[tex]\frac{F_1}{F_2} = 10.125[/tex]
Part b)
Orbital speed is given as
[tex]v = \sqrt{\frac{GM}{r}}[/tex]
so the ratio of two orbital speed is given as
[tex]\frac{v_1}{v_2} = \frac{r_2}{r_1}[/tex]
[tex]\frac{v_1}{v_2} = \sqrt{\frac{4.5r}{r}} = 2.12[/tex]
Part c)
Time period is given as
[tex]T = 2\pi\sqrt{\frac{r^3}{GM}}[/tex]
so the ratio of two time period is given as
[tex]\frac{T_1}{T_2} = \sqrt{\frac{r_1^3}{r_2^3}}[/tex]
[tex]\frac{T_1}{T_2} = \sqrt{\frac{4.5r^3}{r^3}}[/tex]
[tex]\frac{T_1}{T_2} = 9.54[/tex]
The ratios of the gravitational force, orbital velocity, and time period are 10.125, 2.12, and 9.55 respectively.
What is gravitational force?
It is the force exerted by the body to the other body. The force is directly proportional to the product of masses and inversely proportional to the square of the distance between them.
[tex]\rm F = G \dfrac{m_1m_2}{r^2}[/tex]
The more massive planet is 4.5 times as far from the star as the less massive one.
m₁ = 2 m₂
a) The ratio F₁: F₂ of the gravitational forces exerted on the star by the two planets will be
[tex]\rm \dfrac{F_1}{F_2} = \dfrac{m_1r_2^2}{m_2r_1^2}\\\\\\\dfrac{F_1}{F_2} = \dfrac{m_1 (4.5r _1)^2}{2m_1r_1^2}\\\\\\\dfrac{F_1}{F_2} = 10.125[/tex]
b) The ratio v₁: v₂ of the speeds of the two planets will be
Orbital speed is given as
[tex]\rm v = \sqrt{\dfrac{GM}{r}}\\v \propto \dfrac{1}{\sqrt{r}} \\\\ rv^2 \propto c[/tex]
Then
[tex]\rm r_1v_1^2 = r_2v_2^2\\\\\dfrac{v_1}{v_2} = \sqrt{\dfrac{4.5r_1}{r_1}}\\\\\dfrac{v_1}{v_2} = 2.12[/tex]
c) The ratio T₁: T₂ of the orbital periods of the two planets will be
The time period is given as
[tex]\rm T = 2\pi \sqrt{\dfrac{r^3}{GM}}\\\\T \propto \sqrt{r^3}[/tex]
Then
[tex]\rm \dfrac{T_1}{T_2} = \sqrt{\dfrac{r_1^3}{r_2^3}} \\\\\\\dfrac{T_1}{T_2} = \sqrt{\dfrac{(4.5r_1)^3}{r_1}}\\\\\\\dfrac{T_1}{T_2} = 9.55[/tex]
More about the gravitational force link is given below.
https://brainly.com/question/24783651
