Answer
See the detailed answer below.
Work Step by Step
a) According to Kepler's third law the period of a satellite is given by
$$T =\sqrt{\dfrac{4\pi^2r^3}{GM}}=\left(\sqrt{\dfrac{4\pi^2 }{GM}}\right)r^{\frac{3}{2}}$$
where $r$ is the radius of the orbit from the center of the Earth and $M$ is the mass of the Earth.
Hence,
The period of the first satellite is given by
$$T =\left(\sqrt{\dfrac{4\pi^2 }{GM}}\right)r^{\frac{3}{2}}\tag 1$$
And the period of the second satellite is given by
$$T_1=T +\Delta T =\left(\sqrt{\dfrac{4\pi^2 }{GM}}\right)(r+\Delta r)^{\frac{3}{2}}$$
Hence,
$$ \Delta T =\left(\sqrt{\dfrac{4\pi^2 }{GM}}\right)(r+\Delta r)^{\frac{3}{2}}-T $$
Dividing both sides by $T $
$$\dfrac{\Delta T}{T} =\dfrac{\left(\sqrt{\dfrac{4\pi^2 }{GM}}\right)(r+\Delta r)^{\frac{3}{2}}}{T}-1 $$
Plugging $T$ into the second term from (1);
$$\dfrac{\Delta T}{T} =\dfrac{\left(\sqrt{\dfrac{4\pi^2 }{GM}}\right)(r+\Delta r)^{\frac{3}{2}}}{\left(\sqrt{\dfrac{4\pi^2 }{GM}}\right)r^{\frac{3}{2}}}-1 $$
$$\dfrac{\Delta T}{T} =\dfrac{ (r+\Delta r)^{\frac{3}{2}}}{ r^{\frac{3}{2}}}-1 $$
$$\dfrac{\Delta T}{T} =\dfrac{ \left(1+\dfrac{\Delta r}{r}\right)^{\frac{3}{2}}r^{\frac{3}{2}}}{ r^{\frac{3}{2}}}-1 $$
$$\dfrac{\Delta T}{T} = \left(1+\dfrac{\Delta r}{r}\right)^{\frac{3}{2}}- 1\tag 2 $$
Recall that for approximations, $(x\pm 1)^n\approx [1\pm nx]$ when $x\lt\lt 1 $. Applying these since $\Delta r\lt\lt r$.
Thus, $$ \left(1+\dfrac{\Delta r}{r}\right)^{\frac{3}{2}}\approx 1+\frac{3}{2}\dfrac{\Delta r}{r}$$
Plugging into (2);
$$\dfrac{\Delta T}{T} =1+\frac{3}{2}\dfrac{\Delta r}{r} - 1 $$
$$\boxed{\dfrac{\Delta T}{T} = \frac{3}{2}\dfrac{\Delta r}{r} }$$
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b) We know that $$\Delta r=6701-6700=1\;\rm km$$
Plugging into the boxed formula above;
$$\dfrac{\Delta T}{T} = \frac{3}{2}\dfrac{1}{6700} $$
Hence,
$$\dfrac{\Delta T}{T}=\dfrac{3}{13400}$$
where $T$ is the period of the inner satellite.
We need to find the time it takes them to meet again 1 km apart above the Earth which means on the same line from the center of the Earth.
Noting that the reciprocal of $\dfrac{\Delta T}{T}$ is the number of periods needed so they can meet again on the same line.
$$\dfrac{T}{\Delta T}=\dfrac{13400}{3}=4466.67\;\rm times$$
Now we need to find the period of the inner one, which is given by applying Kepler's third law.
$$T = \left(\sqrt{\dfrac{4\pi^2 }{GM}}\right)r^{\frac{3}{2}}$$
$$T = \left(\sqrt{\dfrac{4\pi^2 }{(6.67\times10^{-11})(5.98\times10^{24})}}\right)[6700\times10^3]^{\frac{3}{2}}$$
$$T=5456.05\;rm s$$
Hence, they will meet again after time $t$;
$$\Delta t=4466.67 T=4466.67\times 5456.05=\bf 2.437\times10^7 \rm \;s $$
$$\Delta t=\color{red}{\bf 282.1}\;\rm day$$
