= Solution
{c}
For a satellite of <mass> $m$ moving in the <Newtonian gravitational field> of a much larger, effectively fixed <mass> $M$, put $\mu=GM$. Its conserved <mechanical energy>, with <gravitational potential energy> zero at infinity, is
$$
\boxed{E=\frac12mv^2-\frac{GMm}{r}=-\frac{GMm}{2a}.}
$$
Here $r$ is the instantaneous centre-to-centre <distance>, $v$ the satellite's <speed>, $G$ the <gravitational constant>, and $a$ the <semi-major axis> of its bound <elliptic orbit>. The <kinetic energy> and <potential energy> vary around the <orbit>, while their sum remains constant. Dividing by $m$ and solving for $v$ gives the <vis-viva equation>, $v^2=\mu(2/r-1/a)$.
For this <Hohmann transfer orbit>, the <periapsis> and <apoapsis> <distances> are $r_1=1\,\mathrm{AU}$ and $r_2=1.52\,\mathrm{AU}$, so
$$
a=\frac{r_1+r_2}{2}=1.26\,\mathrm{AU},\qquad
v_1=\sqrt{\frac{\mu}{r_1}}.
$$
The <vis-viva equation> gives the endpoint <speeds>
$$
\boxed{v_{\rm p}=v_1\sqrt{2-\frac1{1.26}}=v_1\sqrt{\frac{76}{63}},\qquad
v_{\rm a}=v_1\sqrt{\frac2{1.52}-\frac1{1.26}}.}
$$
As a check, <conservation of angular momentum> at the two endpoints gives $r_1v_{\rm p}=r_2v_{\rm a}$ because each <velocity> is perpendicular to its radius. Using $1\,\mathrm{AU}=1.49598\times10^{11}\,\mathrm m$ and $\mu=1.32712\times10^{20}\,\mathrm{m^3\,s^{-2}}$ gives $v_1=29.78\,\mathrm{km\,s^{-1}}$ and
$$
\boxed{v_{\rm p}=32.71\,\mathrm{km\,s^{-1}},\qquad
v_{\rm a}=21.52\,\mathrm{km\,s^{-1}}.}
$$
The departure <impulse> must increase the <speed> in the direction of the initial <circular orbit>:
$$
\boxed{\Delta v_1=v_{\rm p}-v_1=2.93\,\mathrm{km\,s^{-1}}.}
$$
These are <Sun>-centred <speeds>; leaving the <Earth>'s own <gravitational field> would require an additional calculation beyond this idealized solar <orbit> transfer.
At $r_2$, the desired <circular orbit> <speed> is $v_2=\sqrt{\mu/r_2}=24.16\,\mathrm{km\,s^{-1}}$. The arrival <impulse> is therefore
$$
\boxed{\Delta v_2=v_2-v_{\rm a}=2.64\,\mathrm{km\,s^{-1}}.}
$$
The required thrust on the spacecraft is forward, tangent to its <orbit>, to accelerate it. The rocket ejects exhaust backwards, opposite to the spacecraft's <velocity>; this distinguishes the direction of thrust from the direction in which the exhaust is fired.
The <Hohmann transfer time> is half the <orbital period> of the transfer <ellipse>. From <Kepler's third law>, relative to the <Earth>'s one-year <orbital period>,
$$
t_H=\frac12(1.26)^{3/2}\,\mathrm{yr}=0.7072\,\mathrm{yr}\simeq258\,\text{days}.
$$
Meanwhile <Mars>'s <orbital period> is $T_M=1.52^{3/2}\,\mathrm{yr}=1.8740\,\mathrm{yr}$. The spacecraft arrives $\pi$ <radians> from its departure <position>. For a rendezvous, <Mars> must reach that same <position> after advancing through $2\pi t_H/T_M$. Thus its initial lead <angle> over the departing spacecraft is
$$
\boxed{\phi_0=\pi-\frac{2\pi t_H}{T_M}
=\pi\left[1-\left(\frac{1.26}{1.52}\right)^{3/2}\right]
=44.15^\circ.}
$$
<Mars> must be ahead by this <angle> in the common direction of <motion> around their <orbits>. At departure the spacecraft and the <Earth> share the inner <orbital position>; arrival occurs on the opposite side of the <Sun>.
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