= Solution
A circular orbit of radius $r$ has speed $v_c(r)=\sqrt{k/r}$. On the transfer <ellipse>, the velocity is tangential at both apsides and has magnitude $h/r$. The first prograde burst therefore changes the speed by
$$
\boxed{\Delta v_1
=\sqrt{\frac{k}{r_1}}
\left(\sqrt{\frac{2r_2}{r_1+r_2}}-1\right)}.
$$
At $r_2$ the satellite must again accelerate in the direction of motion to circularize, by
$$
\boxed{\Delta v_2
=\sqrt{\frac{k}{r_2}}
\left(1-\sqrt{\frac{2r_1}{r_1+r_2}}\right)}.
$$
These are the two impulses of a <Hohmann transfer>.
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