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作者: 大乔治 | 来源:发表于2016-11-28 01:48 被阅读0次

    Exercise 10-Precession of the Perihelion of Mercury (4.10&4.11)

    熊毅恒 2014301020065


    1.abstract

    From the present chapter we will consider several problems that arises in the study of plantery motion.We firstly begin with the simplest situation where a sun and a single planet get moved with interaction.We investigate a few of the properities of this model solar system.The very firts thing that we would like to investigate is the Kepler' laws of motion.Additionally,the inverse-square law and the stability of plantery orbits will be discussed.In the end,we will study the precession of the perihelion of Mercury.

    2.background

    • Kepler's Laws

    According to Newton's law of gravitation the magnitude of gravity is:
    \begin{eqnarray}
    F_G=\frac{GM_S M_E}{r^2}
    \end{eqnarray
    }

    What we aimed to do is to calculate the position of earth as a function of time.From the Newton's Laws of motion we obtain:
    \begin{eqnarray}
    \frac{d2x}{dt2}=\frac{F_{G,x}}{M_E}\quad\frac{d2y}{dt2}=\frac{F_{G,y}}{M_E}
    \end{eqnarray
    }

    thus
    $$F_{G,x}=-\frac{GM_SM_E}{r2}cos\theta=-\frac{GM_SM_Ex}{r2}$$
    We can follow our usual approach and write each of the second-order differential equations as two first-order differential equations. \begin{eqnarray}
    \frac{dv_x}{dt}=-\frac{GM_S x}{r^3} \quad \frac{dx}{dt}=v_x
    \end{eqnarray
    }
    \begin{eqnarray}
    \frac{dv_y}{dt}=-\frac{GM_S y}{r^3} \quad \frac{dy}{dt}=v_y
    \end{eqnarray
    }
    We can use Euler-Cromer method to solve the equations: $$r_i=(x_i2+y_i2)^{1/2}$$
    $$ v_{x,i+1}=v_{x,i}-\frac{4\pi{2}x_i}{r_i{3}}\Delta t\qquad x_{i+1}=x_i+v_{x,i+1}\Delta t$$
    $$v_{y,i+1}=v_{y,i}-\frac{4\pi{2}y_i}{r_i{3}}\Delta t\qquad y_{i+1}=y_i+v_{y,i+1}\Delta t$$
    The orbital trajectory for a body of reduced mass $\mu$ is given in polar coordinates by $$ \frac{d2}{d\theta2}(\frac{1}{r})+\frac{1}{r}=-\frac{\mu r2}{L2}F(r)$$

    Also the momentum L is a constant,
    Thus we have
    $$v_{max}=\sqrt{GM_S}\sqrt{\frac{1+e}{a(1-e)}(1+\frac{M_P}{M_S})}$$
    $$v_{min}=\sqrt{GM_S}\sqrt{\frac{1-e}{a(1+e)}(1+\frac{M_P}{M_S})}$$

    • The inverse-square law and the stability of plantery orbits

    Suppose that the force law deviates slightly from an inverse-square dependence.To be specific,suppose that the gravational force is of the form:
    $$F_G=\frac{GM_SM_E}{r^{\beta}}$$

    On the condition of $\beta=2$ we obtain the inverse-square law.Nevertheless we also want to conisder the motion of our planet for values of $\beta$ not equal to 2.The elleptical orbits with this force law can be simulated with the planteary motion program given above by simply changing the exponent of r in the equations for the velocity.

    • The precession of the perihelion of Mercury

    In fact, there are deviations from the laws we have discussed.These deviations come from a number of sources, including the effects of the planets on each other. It turns out that this is a problem for which very few exact results are known.
    Of most importance is the precession of the perihelion of Mercury. The planets whose orbits deviate the most from circular are Mercury and Pluto.
    All we have to do is simulate the orbital motion using the force law predicted by the general relativity and measure the rate of precession of the orbit.

    \begin{eqnarray}
    F_G\approx\frac{GM_S M_M}{r2}(1+\frac{\alpha}{r2})
    \end{eqnarray
    }
    $$-\frac{GM_SM_M}{r_1}+\frac{1}{2}M_Mv_12=-\frac{GM_SM_M}{r_2}+\frac{1}{2}M_Mv_22$$
    ( $b=a\sqrt{1-e^2}$ and $r_1v_1=bv_2$ )

    3.Mainbody

    1. The influence of eccentricities
    • First we can see the normal movement trajectory of a Mercury orbiting Sun.
    • change the eccentricities from 0.206 to 0.306, 0.406 and 0.506


      e=0.306
      e=0.406 .png
      e=0.606

      We shall see with bigger eccentricity, the trajectory becomes more elliptical, as suggested by the definition of eccentricity.

    • By giving different duration of time from T=1 to T=5 and 20 , It will be more evidently that the trajectory is gradually turning the orientation as predicted.(when e=0.406)

    T=1
    T=5
    T=20
    1. The influence of initial velocity
      We set different initial velocity to investigate the behavior of Mercury. As the increase of the Mercury will at last went away from the solar system with the second universal velocity.



      3.The influence of $\beta$
      We change the value of $\beta$ and shall see that except the situation where $\beta=0$ ,the trajectory is not closed.

    beta=2.0
    beta=2.2
    beta=2.4
    beta=2.6

    4.The influence of $\alpha$
    Having changed the value of $\alpha$ we observe some misbehave of the curves.


    Let us return to the precession of the perihelion of Mercury.We shall see the slope of the lines remains constant as predicted.

    4.conlusion

    With the help of Kepler's Laws and the Euler-Cromer method, we can solve the motion of the simplest situation where a sun and a single planet get moved with interaction. We investigate a few of the properities of this model solar system. Then we can add the inverse-square law and the stability of plantery orbits will be discussed. Thus e can learn the precession of the perihelion of Mercury. And we can try to plot the figure of the precession of perihelion of Mercury. Thus we can know the precession of perihelion of Mercury more clear.

    acknowledgement

    The Lesson plan Chapter 3 of Cai Hao
    zt

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