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

Question

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 .

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Edana Edana 4 years 2021-08-02T07:39:33+00:00 1 Answers 1268 views 0

Answers ( )

  1. RuslanHeatt
    1
    2021-08-02T07:41:14+00:00
    Reply

    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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