Answer
See the detailed answer below.
Work Step by Step
a) We know that the energy and the momentum are conserved.
So,
$$E_1=E_2$$
$$K_1+U_{g1}=K_2+U_{g2}$$
$$\frac{1}{2}\color{red}{\bf\not}
mv_1^2+\dfrac{-GM_E\color{red}{\bf\not}
m}{r_1 }=\frac{1}{2}\color{red}{\bf\not}
mv_2^2+\dfrac{-GM_E\color{red}{\bf\not}
m}{r_2 }$$
Multiplying by 2;
$$ v_1^2+\dfrac{-2GM_E }{r_1}= v_2^2+\dfrac{-2GM_E }{r_2}\tag 1$$
According to the conservation of angular momentum:
$$L_i=L_f$$
$$ \color{red}{\bf\not}mv_1r_1 \color{red}{\bf\not}\sin90^\circ= \color{red}{\bf\not}mv_2r_2 \color{red}{\bf\not}\sin90^\circ$$
$$v_1r_1=v_2r_2$$
So,
$$v_1=\dfrac{v_2r_2}{r_1}\tag 2$$
Plugging into (1);
$$ \left(\dfrac{v_2r_2}{r_1}\right)^2+\dfrac{-2GM_E }{r_1 }= v_2^2+\dfrac{-2GM_E }{r_2 } $$
$$ \dfrac{v_2^2r_2^2}{r_1^2,}-v_2^2= \dfrac{-2GM_E }{r_2 } +\dfrac{2GM_E }{r_1 }$$
$$ v_2^2\left[\dfrac{r_2^2}{r_1^2}-1\right]=-2GM_E\left[ \dfrac{ 1}{r_2 } -\dfrac{1}{r_1 }\right]$$
$$ v_2^2\left[\dfrac{r_2^2-r_1^2}{r_1^{ \color{red}{\bf\not}2}} \right]= -2GM_E\left[ \dfrac{ r_1 -r_2 }{ \color{red}{\bf\not}r_1 r_2 } \right]$$
$$ v_2^2\left[\dfrac{ \color{red}{\bf\not}(r_2-r_1)(r_2+r_1)}{r_1 } \right]= 2GM_E\left[ \dfrac{ \color{red}{\bf\not}(r_2 -r_1) }{ r_2 } \right]$$
$$v_2^2=\dfrac{2GM_E}{r_2}\cdot \dfrac{r_1}{r_1+r_2}$$
Thus,
$$v_2=\sqrt{\dfrac{2GM_E}{r_2}\cdot \dfrac{r_1}{r_1+r_2}}$$
$$\boxed{v_2'=\sqrt{\dfrac{2GM (r_1/r_2)}{r_1+r_2}}}$$
Plugging into (2);
$$v_1=\sqrt{\dfrac{2GM_E(r_1/r_2)}{r_1+r_2}}\dfrac{ r_2}{r_1}=\sqrt{\dfrac{2GM_E( \color{red}{\bf\not}r_1/ \color{red}{\bf\not}r_2)}{r_1+r_2}\cdot \dfrac{ r_2^{ \color{red}{\bf\not}2}}{r_1^{ \color{red}{\bf\not}2}}}$$
$$\boxed{v_1'=\sqrt{\dfrac{2GM (r_2/r_1)}{r_1+r_2}}}$$
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b) We know that
$$r_1=R_E+(300\times 10^3)=(6.37\times 10^6)+(300\times 10^3)$$
$$r_1=\bf 6.67\times 10^6\;\rm m$$
and
$$r_2=R_E+(35,900\times 10^3)=(6.37\times 10^6)+(35,900\times 10^3)$$
$$r_2=\bf 4.227\times10^7\;\rm m$$
To find $v_1'$ we can use the second boxed formula above:
$$v_1'=\sqrt{\dfrac{2GM (r_2/r_1)}{r_1+r_2}}$$
Plugging the known;
$$v_1'=\sqrt{\dfrac{2(6.67\times 10^{-11})(5.98\times 10^{24}) ([4.227\times10^7]/[6.67\times 10^6])}{(6.67\times 10^6)+(4.227\times10^7)}}$$
$$v_1'=\color{red}{\bf 1.0164\times 10^4}\;\rm m/s$$
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c) We can use the work-kinetic energy theorem to find the work done by the rocket's motor.
$$W_1=\Delta K=\frac{1}{2}m{v_1^2}'-\frac{1}{2}m{v_1^2} $$
We found $v_1'$, so we need to find the speed of the circular orbit $v_1$ when the satellite was at 300 km above the ground.
Using Newton's second law:
$$\dfrac{GM\color{red}{\bf\not}m}{r_1^{ \color{red}{\bf\not}2}}=\dfrac{ \color{red}{\bf\not}mv_1^2}{ \color{red}{\bf\not}r_1}$$
Hence,
$$v_1=\sqrt{\dfrac{GM}{r_1}}=\sqrt{\dfrac{(6.67\times 10^{-11})(5.98\times 10^{24})}{6.67\times 10^6}}$$
$$v_1 =\color{blue}{\bf 7.733\times 10^3}\;\rm m/s$$
Plugging the two velocities into the work formula above:
$$W_1= \frac{1}{2}m\left[{v_1^2}'- {v_1^2} \right]$$
$$W_1= \frac{1}{2}(1000)\left[(1.0164\times 10^4)^2- (7.733 \times 10^3)^2 \right]$$
$$W_1=\color{red}{\bf 2.17\times10^{10}}\;\rm J$$
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d)
Using the boxed formula above for $v_2'$:
$$ v_2'=\sqrt{\dfrac{2GM (r_1/r_2)}{r_1+r_2}} $$
$$v_2'=\sqrt{\dfrac{2(6.67\times 10^{-11})(5.98\times 10^{24}) ([6.67\times 10^6]/[4.227\times10^7])}{(6.67\times 10^6)+(4.227\times10^7)}}$$
$$v_2'=\color{red}{\bf 1.604\times 10^3}\;\rm m/s$$
Now we need to find $v_2$ for circular orbit at the same distance:
$$v_2=\sqrt{\dfrac{GM}{r_2}}=\sqrt{\dfrac{(6.67\times 10^{-11})(5.98\times 10^{24})}{4.227\times 10^7}}$$
$$v_2 =\color{red}{\bf 3.072\times 10^3}\;\rm m/s$$
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e) By the same approach as part (c);
$$W_2= \frac{1}{2}m\left[{v_2^2}- {v_2^2}'\right]$$
$$W_2= \frac{1}{2}(1000)\left[(3.072\times 10^3)^2- ( 1.604\times 10^3)^2 \right]$$
$$W_2=\color{red}{\bf 3.432\times10^9}\;\rm J$$
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f) The total work done is given by
$$W_{tot}=W_1+W_2$$
Plugging from above;
$$W_{tot}=(2.17\times10^{10})+(3.432\times10^9)$$
$$W_{tot}=\color{red}{\bf 2.51\times10^{10}}\;\rm J$$
