Three different planet-star systems, which are far apart from one another, are shown above. The masses of the planets are much less than the masses of the stars.

In System A , Planet A of mass Mp orbits Star A of mass Ms in a circular orbit of radius R .

In System B , Planet B of mass 4Mp orbits Star B of mass Ms in a circular orbit of radius R .

In System C , Planet C of mass Mp orbits Star C of mass 4Ms in a circular orbit of radius R .

(a) The gravitational force exerted on Planet A by Star A has a magnitude of F0 . Determine the magnitudes of the gravitational forces exerted in System B and System C .

___ Magnitude of gravitational force exerted on Planet B by Star B

___ Magnitude of gravitational force exerted on Planet C by Star C

(b) How do the tangential speeds of planets B and C compare to that of Planet A ? In a clear, coherent paragraph-length response that may also contain equations and/or drawings, provide claims about

why the tangential speed of Planet B is either greater than, less than, or the same as that of Planet A , and

why the tangential speed of Planet C is either greater than, less than, or the same as that of Planet A .

a) 4F0

b) Speed of planet B is the same as speed of planet A

Speed of planet C is twice the speed of planet A

Explanation:a)

The magnitude of the gravitational force between two objects is given by the formula

[tex]F=G\frac{m_1 m_2}{r^2}[/tex]

where

G is the gravitational constant

m1, m2 are the masses of the 2 objects

r is the separation between the objects

For the system planet A – Star A, we have:

[tex]m_1=M_p\\m_2 = M_s\\r=R[/tex]

So the force is

[tex]F_A=G\frac{M_p M_s}{R^2}=F_0[/tex]

For the system planet B – Star B, we have:

[tex]m_1 = 4 M_p\\m_2 = M_s\\r=R[/tex]

So the force is

[tex]F=G\frac{4M_p M_s}{R^2}=4F_0[/tex]

So, the magnitude of the gravitational force exerted on planet B by star B is 4F0.

For the system planet C – Star C, we have:

[tex]m_1 = M_p\\m_2 = 4M_s\\r=R[/tex]

So the force is

[tex]F=G\frac{M_p (4M_s)}{R^2}=4F_0[/tex]

So, the magnitude of the gravitational force exerted on planet C by star C is 4F0.

b)

The gravitational force on the planet orbiting around the star is equal to the centripetal force, therefore we can write:

[tex]G\frac{mM}{r^2}=m\frac{v^2}{r}[/tex]

where

m is the mass of the planet

M is the mass of the star

v is the tangential speed

We can re-arrange the equation solving for v, and we find an expression for the speed:

[tex]v=\sqrt{\frac{GM}{r}}[/tex]

For System A,

[tex]M=M_s\\r=R[/tex]

So the tangential speed is

[tex]v_A=\sqrt{\frac{GM_s}{R}}[/tex]

For system B,

[tex]M=M_s\\r=R[/tex]

So the tangential speed is

[tex]v_B=\sqrt{\frac{GM_s}{R}}=v_A[/tex]

So, the speed of planet B is the same as planet A.

For system C,

[tex]M=4M_s\\r=R[/tex]

So the tangential speed is

[tex]v_C=\sqrt{\frac{G(4M_s)}{R}}=2(\sqrt{\frac{GM_s}{R}})=2v_A[/tex]

So, the speed of planet C is twice the speed of planet A.