Drops of Jupiter

From the previous problem, we recall that $\frac{dr}{dt} \; = \; -\sqrt{\frac{2GM (R - r)}{Rr}}$ is the upwards velocity of a particle at height $r$ above the centre of a planet of mass $M$ , which starts from rest at height $R$ . Separating variables and solving this differential equation tells us that $t \; = \; \int_r^R \sqrt{\frac{Rr}{2GM(R-r)}}\,dr$ is the time it takes this particle to reach the height $r$ above the centre of the planet. Making the single substitution $r = R\sin^2\theta$ yields $\begin{array}{rcl} t &= & \displaystyle \int_{\sin^{-1}\sqrt{\frac{r}{R}}}^{\frac12\pi} \sqrt{\frac{R^2\sin^2\theta}{2GM R\cos^2\theta}}\,2R\sin\theta \cos\theta\,d\theta \\ & = & \displaystyle \sqrt{\frac{2R^3}{GM}}\int_{\sin^{-1}\sqrt{\frac{r}{R}}}^{\frac12\pi} \sin ^2\theta\,d\theta \; =\; \sqrt{\frac{R^3}{2GM}}\int_{\sin^{-1}\sqrt{\frac{r}{R}}}^{\frac12\pi} (1 - \cos2\theta)\,d\theta \\ & = & \displaystyle \sqrt{\frac{R^3}{2GM}}\Big[\theta - \sin\theta\cos\theta\Big]_{\sin^{-1}\sqrt{\frac{r}{R}}}^{\frac12\pi} \; = \; \sqrt{\frac{R^3}{2GM}}\Big[\tfrac12\pi - \sin^{-1}\sqrt{\tfrac{r}{R}} + \tfrac{1}{R}\sqrt{r(R-r)}\Big] \\ & = & \displaystyle \sqrt{\frac{R^3}{2GM}}\Big[\cos^{-1}\sqrt{\tfrac{r}{R}} + \tfrac{1}{R}\sqrt{r(R-r)}\Big] \end{array}$ Substituting the given values for the various parameters gives a time of $\boxed{36.1}$ days.

$G$ is the gravitational constant $6.674 \cdot 10^{-11} \, m^{3} kg^{-1} s^{-2}$

$M$ is the mass of the planet $1.9 \cdot 10^{27}kg$

$g$ is acceleration due to gravity, the gravitational field

$L$ is the initial distance of the object from the center of the planet $10^{10}m$

$r$ is the distance of the object from the center of the planet at any given time

$v$ is the speed of the object

.

Acceleration g is the time-derivative of velocity v . This acceleration is represented with negative numbers because its vector points in the direction opposite that of the position vector r .

$\frac{dv}{dt} = g = - \frac{GM}{r^{2}}$

$\frac{dv}{dt} = \frac{dv}{dr} \cdot \frac{dr}{dt} = \frac{dv}{dr} \cdot v = -GMr^{-2}$

$v \: dv = -GMr^{-2} \, dr$ , $\; \; \int_{v(a)}^{v(b)} v \: dv = -GM \int_a^b r^{-2} \, dr$

The initial position is $a = L$ , and the initial velocity is $v(a) = 0$ . Instead of using dummy variables I'm going to re-use $v$ for $v(b)$ and re-use $r$ for $b$ .

$\int_0^v v \: dv = \frac{1}{2} v^{2}$

$-GM \int_L^r r^{-2} \, dr = GMr^{-1}|_L^{r} = GM(r^{-1} - L^{-1})$

$GM(r^{-1} - L^{-1}) = \frac{1}{2} v^{2}$ , $\; \; v^{2} = 2GM(r^{-1} - L^{-1})$

$v = -\sqrt{2GM(r^{-1} - L^{-1})}$

The velocity needs to be represented by negative numbers because its vector is pointing in the direction opposite that of the position vector. If I don't use a negative velocity I will ended up with a negative number for the time.

$\frac{dr}{dt} = v$ , $\; \; dr = v \, dt$ , $\; \; dt = v^{-1} \, dr$

$\large dt = -[2GM(r^{-1} - L^{-1})]^{-1/2} \, dr$

I'm going to use a trigonometric substitution, but first I need to re-write $(r^{-1} - L^{-1})$ as $(u^{2} - a^{2})$ , so I make these substitutions

$u = r^{-1/2}$ , $\; \; a = L^{-1/2}$ , $\; \; dr = -2u^{-3} \, du$

Now for the trig substitutions draw a right triangle with an angle $\theta$ . Label the hypotenuse $u$ . Label the side adjacent to $\theta$ as $a$ . Then the other side is $\sqrt{u^{2} -a^{2}}$ . From this triangle you can find the below useful relations.

$cot(\theta) = a(u^{2}-a^{2})^{-1/2}$ , $\; \; cos(\theta) = au^{-1}$ , $\; \; -sin(\theta) \, d \theta = -au^{-2} \, du$

$du = a \; tan(\theta) \, sec(\theta) \, d \theta$

Now to start putting it together

$(r^{-1} - L^{-1})^{-1/2} \, dr = a^{-1} \; cot(\theta) \cdot [-2a^{-3} \; cos^{3}(\theta)] \cdot [a \; tan(\theta) \, sec(\theta)] \, d \theta =$

$= -2a^{-3} \; cos^{2}(\theta) \, d \theta = -a^{-3} \; [1 + cos(2\theta)] \, d \theta$

Getting close now. Note that $acos(\phi)$ is the inverse cosine, also known as the arc-cosine

$\int (r^{-1} - L^{-1})^{-1/2} \, dr = -a^{-3} \: [ \int d \theta + \int cos(2\theta) \, d \theta ] = -a^{-3} \; [\theta + sin(\theta) \, cos(\theta)] + C =$

$= -a^{-3} \: [acos(au^{-1}) + a \, \sqrt{u^{2}-a^{2}} \cdot u^{-2}] +C =$

$= -a^{-3} \: [acos(L^{-1/2} \cdot r^{1/2}) + L^{-1/2} \sqrt{r^{-1} - L^{-1}} \cdot r] + C =$

$= -L^{3/2} \: [acos( \sqrt{r/L}) + L^{-1/2} \cdot r \, \sqrt{r^{-1} - L^{-1}}] + C =$

$= -L^{3/2} \: acos( \sqrt{r/L}) - rL \: \sqrt{r^{-1} - L^{-1}} + C$

Now to find the time

$t = -(2GM)^{-1/2} \: \int (r^{-1} - L^{-1})^{-1/2} \, dr$

When $r = L$ then $t = C = 0$ .

$\large t = (2GM)^{-1/2} \: \Bigg( \, L^{3/2} \: acos \Big( \sqrt{r/L} \Big) + rL \: \sqrt{r^{-1} - L^{-1}} \, \Bigg)$

Setting $r$ equal to $10^{8}m$ you get $3.12 \cdot 10^{6}s$ , or $36.1$ days.