Dynamics | Part IA, 2008

A body of mass mm moves in the gravitational field of a much larger spherical object of mass MM located at the origin. Starting from the equations of motion

r¨rθ˙2=GMr2,rθ¨+2r˙θ˙=0,\begin{aligned} \ddot{r}-r \dot{\theta}^{2} &=-\frac{G M}{r^{2}}, \\ r \ddot{\theta}+2 \dot{r} \dot{\theta} &=0, \end{aligned}

show that:

(i) the body moves in an orbit of the form

h2uGM=1+ecos(θθ0)\frac{h^{2} u}{G M}=1+e \cos \left(\theta-\theta_{0}\right)

where u=1/r,hu=1 / r, h is the constant angular momentum per unit mass, and ee and θ0\theta_{0} are constants;

(ii) the total energy of the body is

E=mG2M22h2(e21)E=\frac{m G^{2} M^{2}}{2 h^{2}}\left(e^{2}-1\right)

A meteorite is moving very far from the Earth with speed VV, and in the absence of the effect of the Earth's gravitational field would miss the Earth by a shortest distance bb (measured from the Earth's centre). Show that in the subsequent motion

h=bVh=b V


e=[1+b2V4G2M2]12e=\left[1+\frac{b^{2} V^{4}}{G^{2} M^{2}}\right]^{\frac{1}{2}}

Use equation ()(*) to find the distance of closest approach, and show that the meteorite will collide with the Earth if

b<[R2+2GMRV2]12b<\left[R^{2}+\frac{2 G M R}{V^{2}}\right]^{\frac{1}{2}}

where RR is the radius of the Earth.

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