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Sergeeva-Olga [200]
3 years ago
15

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

ess 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 .
Physics
1 answer:
alex41 [277]3 years ago
3 0

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

F=G\frac{m_1 m_2}{r^2}

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:

m_1=M_p\\m_2 = M_s\\r=R

So the force is

F_A=G\frac{M_p M_s}{R^2}=F_0

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

m_1 = 4 M_p\\m_2 = M_s\\r=R

So the force is

F=G\frac{4M_p M_s}{R^2}=4F_0

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:

m_1 = M_p\\m_2 = 4M_s\\r=R

So the force is

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

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:

G\frac{mM}{r^2}=m\frac{v^2}{r}

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:

v=\sqrt{\frac{GM}{r}}

For System A,

M=M_s\\r=R

So the tangential speed is

v_A=\sqrt{\frac{GM_s}{R}}

For system B,

M=M_s\\r=R

So the tangential speed is

v_B=\sqrt{\frac{GM_s}{R}}=v_A

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

For system C,

M=4M_s\\r=R

So the tangential speed is

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

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

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