Falling Towards A Gas Giant

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

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

$\frac{dv}{dr} \cdot v = -\frac{GM}{r^2}$

$\int_{0}^{v^*} vdv = \int_{10^{10}}^{10^8} -\frac{GM}{r^2} dr$

$\frac{v^2}{2} \big|_{0}^{v^*} = \frac{GM}{r} \big|_{10^{10}}^{10^8}$

$\frac{v^{*2}}{2} = GM(10^{-8} - 10^{-10})$

$v^* = \sqrt{2 \cdot 6.674 \cdot 10^{-11} \cdot 1.9 \cdot 10^{27} \cdot (10^{-8} - 10^{-10}) }$

$v^* = \fbox{50107.47}$

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

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

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

$m$ is the mass of the object

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

$K$ is the kinetic energy of the object

$v$ is the speed of the object

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Since gravity is the only force acting on the object, the kinetic energy of the object at any position is equal to the work done on the object by gravity in moving it to that position. The gravitational force acting on the object is

$F = GMm/r^{2}$

The gravity varies greatly over the distance L, so you have to go back to the definition of work. As an object moves along a path S the work done on the object by a force F is the integral over S of the scalar product $F \cdot dx$ , where dx is the differential of the displacement vector.

$W = \int_S F \cdot dx$

In our case the gravity and displacement vectors point in the same direction so the scalar product is simply F dx, and dx = -dr. So we can write work as

$W = -\int_L^r GMmr^{-2} \, dr = GMmr^{-1}|_L^r = GMm(r^{-1} - L^{-1}) = K$

Instead of using a dummy variable I just re-used $r$ . I integrated from position $L$ down to a lower position $r$ . Now I have the kinetic energy K. I also know kinetic energy is one-half the mass times the square of the speed.

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

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

When $r$ is set equal to $10^{8}m$ you get

$v = 50,107.47 \:m/s$