Answer
b) 15.3 days
c) The acceleration increases since the distance between the astronauts decreases.
Work Step by Step
b)
Consider the two astronauts as shown in the figure. We refer to them as A and B, with $m_A = 65\mathrm{kg}$ and $m_B = 72\mathrm{kg}$. They are both attracted to each other due to gravity, with the force on A($F$), being equal and opposite to the force on B ($-F$).
Here the distance between A and B is $S=20\mathrm{m}$.
The gravitational force on A is
$F_A = F = \frac{Gm_Am_B}{S^2} = \frac{6.67\times10^{-11} \mathrm{m^3 kg^{-1}s^{-2}}\times 65\mathrm{kg}\times 72\mathrm{kg}}{(20\mathrm{m})^2} = 7.80\times 10^{-10} N $
The gravitational force on B is
$F_B=-F = -\frac{Gm_Am_B}{S^2} = \frac{6.67\times10^{-11} \mathrm{m^3 kg^{-1}s^{-2}}\times 65\mathrm{kg}\times 72\mathrm{kg}}{(20\mathrm{m})^2} =- 7.80\times 10^{-10} N $
Assuming the accelerations to be constant, the acceleration of A is
$a_A = F_A/m_A =\frac{7.80\times 10^{-10} N}{65\mathrm{kg}} =1.2\times 10^{-11} \mathrm{ms^{-2}}$.
Acceleration of B is
$a_B = F_B/m_B =-\frac{7.80\times 10^{-10} N}{72\mathrm{kg}} =-1.08\times 10^{-11} \mathrm{ms^{-2}}$.
The acceleration of A relative to B is
$a_{AB} = a_A-a_B = 1.2\times 10^{-11} \mathrm{ms^{-2}}-(-1.08\times 10^{-11} \mathrm{ms^{-2}})= 2.28\times 10^{-11} \mathrm{ms^{-2}}$.
From the kinematic equations, $S = 0.5a_{AB}t^2 \implies t = \sqrt{2r/a_{AB}}$
$t = \sqrt{\frac{2\times20\mathrm{m}}{2.28\times 10^{-11} \mathrm{ms^{-2}}}} = 1.32\times10^6\mathrm{s} = 15.3 \mathrm{days} $.
Thus the astronauts will take $15.3\mathrm{days}$ to get to each other if the acceleration remained constant.
c) The acceleration, however, does not remain constant. As the astronauts get closer, the force due to gravity and hence their acceleration increases since $F\propto S^{-2}$ and $S$ decreases. Thus, the time taken will be less than 15.3 days.
