Question

Two astronauts have deserted their spaceship in a region of space far from the gravitational attraction of any other body. Each has a mass of $100$kg and they are $100$m apart. They are initially at rest relative to one another. How long will it be before the gravitational attraction brings them $1$cm closer together?

• A. $2.52$ days
• B. $1.41$ days
• C. $0.70$ days
• D. $1.41$s

Answer 1

The correct option is B $1.41$ days

$a_1=$ acceleration of first
$=\frac{G{m}_1{m}_2}{r_2}\times \frac{1}{m_1}=\frac{G{m}_2}{r^2}=\frac{6.67\times {10}^{-11}\times 100}{(100{)}^2}$
$=6.67\times {10}^{-13}m{s}^{-1}$
$a_2$= acceleration of second
$=\frac{G{m}_1}{r^2}=\frac{6.67\times {10}^{-11}\times 100}{(100{)}^2}=6.67\times {10}^{-13}m{s}^{-1}$
Net acceleration of approach $a=a_1+a_2=2\times 6.67\times {10}^{-13}m{s}^{-1}$Now, $s=\frac{1}{2}a{t}^2$
$1\times {10}^{-2}=\frac{1}{2}\times 6.67\times {10}^{-13}\times 2\times t^2$Solving, we get $t=1.41$ days.