Answer
$42\;\rm m$
Work Step by Step
According to Newton's third law, the force exerted by the astronaut on the satellite is equal in magnitude but opposite in direction to the force exerted by the satellite on the astronaut.
Now we need to find the acceleration of each object during the first half second of applying force to find their final speed after this small time interval.
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Let's start with the astronaut:
$$\sum F_{\rm on\;A}=m_{\rm A}a_{\rm A}=100$$
Thus, the acceleration of the astronaut is
$$a_{\rm A}=\dfrac{100}{m_{\rm A}}=\dfrac{100}{80}=\color{blue}{\bf 1.25}\;\rm m/s^2$$
We know that he starts to move from rest, so the distance traveled during this acceleration time is
$$\Delta x_{\rm A,1}=v_it+\frac{1}{2}at_1^2=0+\frac{1}{2}\cdot 1.25\cdot 0.5^2=\dfrac{5}{32}\;\rm m$$
His final speed after the time of acceleration is given by
$$v_f=v_i+at=0+1.25\cdot 0.5=\dfrac{5}{8}\;\rm m/s$$
Thus the distance traveled during the constant speed time interval of $\rm 1\;minute-0.5 \;second=59.5\;seconds$ is given by
$$\Delta x_{\rm A,2}=vt=\dfrac{5}{8}\cdot 59.5=\dfrac{595}{16}\;\rm m$$
Therefore,
$$\Delta x_{\rm A,tot}=\dfrac{5}{32}+\dfrac{595}{16}=\color{red}{\bf 37.3}\;\rm m$$
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Let's work on the satellite :
$$\sum F_{\rm on\;S}=m_{\rm S}a_{\rm S}=-100$$
the negative sign is due to the opposite direction
Thus, the acceleration of the satellite is
$$a_{\rm S}=-\dfrac{100}{m_{\rm S}}=\dfrac{-100}{640}=\color{blue}{\bf-0.156}\;\rm m/s^2$$
We know that it starts to move from rest, so the distance traveled during this acceleration time is
$$\Delta x_{\rm S,1}=v_it+\frac{1}{2}at_1^2=0-\frac{1}{2}\cdot 0.156\cdot 0.5^2=\dfrac{-39}{2000}\;\rm m$$
Its final speed after the time of acceleration is given by
$$v_f=v_i+at=0-0.156\cdot 0.5=\dfrac{-39}{500}\;\rm m/s$$
Thus the distance traveled during the constant speed time interval of $\rm 1\;minute-0.5 \;second=59.5\;seconds$ is given by
$$\Delta x_{\rm S,2}=vt=\dfrac{-39}{500}\cdot 59.5=\dfrac{-4641}{1000}\;\rm m$$
Therefore,
$$\Delta x_{\rm S,tot}=\dfrac{-39}{500}+\dfrac{-4641}{1000}=\color{red}{\bf -4.72}\;\rm m$$
Therefore, the separation distance is given by
$$d=x_{\rm A}-x_{\rm S}=37.3-(-4.72)=\color{magenta}{\bf 42}\;\rm m$$
