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Thread: Differential Equation: Newton's Law of Gravitation

  1. #1
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    Differential Equation: Newton's Law of Gravitation

    This is a problem we were given to practice differential equations and I have not the darndest clue of what to do. Can someone walk me through it so I can practice on my own.

    According to Newton’s Law of Universal Gravitation, the gravitational force on an object of mass m that has been projected vertically upward from Earth’s surface is

    F = (mgR^2) / ((x+R)^2)

    where x = x(t) is the object’s distance above the surface at time t, R is Earth’s radius, and g is the acceleration due to gravity. Also, by Newton’s Second Law, F = ma = m( dv/dt ) and so

    m(dv/dt) = (−mgR^2) / ((x+R)^2)

    a) Suppose a rocket is fired vertically upward with an initial velocity v0. Let h be the maximum height above the surface reached by the object. Show that

    v0 = sqrt((2gRh)/(R+h))

    (Hint: by the chain rule, m(dv/dt) = mv(dv/dx).)

    b) Calculate Ve = lim(as h goes to infinity) V0. This limit is called the escape velocity for Earth.

    c) Use R = 3960mi and g = 32ft/sec^2 to calculate Ve in feet per second and in miles per second.
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    Quote Originally Posted by TreeMoney View Post
    a) Suppose a rocket is fired vertically upward with an initial velocity v0. Let h be the maximum height above the surface reached by the object. Show that

    v0 = sqrt((2gRh)/(R+h))

    (Hint: by the chain rule, m(dv/dt) = mv(dv/dx).)
    We have:
    $\displaystyle mv \frac{dv}{dx} = - \frac{mgR^2}{(R + x)^2}$

    $\displaystyle v \frac{dv}{dx} = - \frac{gR^2}{(R+x)^2}$

    $\displaystyle v dv = - \frac{gR^2}{(R+x)^2} dx$

    Now, at the surface $\displaystyle v = v_0$ and x = 0. At the maximum height v = 0 and x = h, so:

    $\displaystyle \int_{v_0}^0 v \, dv = - \int_0^h \frac{gR^2}{(R+x)^2}dx$

    The LHS is:
    $\displaystyle \frac{1}{2}v^2 |_{v_0}^0 = - \frac{1}{2}v_0^2$

    For the RHS, let y = R + x:
    $\displaystyle - \int_0^h \frac{gR^2}{(R+x)^2}dx = - \int_R^{h+R} \frac{gR^2}{y^2}$ = $\displaystyle -gR^2 \cdot \frac{-1}{y} |_R^{h+R}$ = $\displaystyle gR^2 \left ( \frac{1}{h+R} - \frac{1}{R} \right )$ = $\displaystyle - \frac{gRh}{h+R}$

    Equating the LHS and the RHS:
    $\displaystyle - \frac{1}{2}v_0^2 = - \frac{gRh}{h+R}$

    $\displaystyle v_0^2 = \frac{2gRh}{h+R}$

    $\displaystyle v_0 = \sqrt{ \frac{2gRh}{h+R}}$

    -Dan
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    Quote Originally Posted by TreeMoney View Post
    b) Calculate Ve = lim(as h goes to infinity) V0. This limit is called the escape velocity for Earth.
    From a) we have that:

    $\displaystyle v_0 = \sqrt{ \frac{2gRh}{h+R}}$

    The escape velocity (speed, really) is the speed of launch required such that the maximum "height" of the trajectory is infinite. (So v = 0 at h = infinity.)

    So
    $\displaystyle v_e = \lim_{h \to \infty}\sqrt{ \frac{2gRh}{h+R}}$

    $\displaystyle v_e = \lim_{h \to \infty}\sqrt{ \frac{2gRh}{h+R} \cdot \frac{ \frac{1}{h}}{\frac{1}{h}}} = \lim_{h \to \infty}\sqrt{ \frac{2gR}{1+\frac{R}{h}}}$

    $\displaystyle v_e = \sqrt{2gR}$

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by TreeMoney View Post
    c) Use R = 3960mi and g = 32ft/sec^2 to calculate Ve in feet per second and in miles per second.
    $\displaystyle v_e = \sqrt{2gR}$

    $\displaystyle R = 3960 \, mi = \frac{3960 \, mi}{1} \cdot \frac{5280 \, ft}{1 \, mi} = 20908800 \, ft$

    So
    $\displaystyle v_e = \sqrt{2 \cdot 32 \cdot 20908800} = 36580.9 \, ft/s$

    and

    $\displaystyle 36580.9 \, ft/s = \frac{36580.9 \, ft}{1 \, s} \cdot {1 \, mi}{5280 \, ft} = 6.9282 \, mi/s$

    or $\displaystyle v_e = 6.923 \, mi/s$ to 3 significant figures.

    -Dan
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    Thanks!!!!

    Thanks Dan, It's much appreciated!!!!
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