Dealing with gravitational force

Classical Mechanics Level pending

Two spherical bodies of masses M and 5M and radii R and 2R are released in free space with initial separation between their centres equal to 12R . If they attract each other due to gravitational force only, then the distance covered by the smaller body before collision is

1.5R 7.5R 4.5R 2.5R

Debanik Samanta by Debanik Samanta , 20, India

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1 solution

I saw no way out of using a numerical solution to set of interrelated differential equations.

y1 ′ ′ ( t ) = a1 ( y5 ( t ) − y1 ( t ) ) , y5 ′ ′ ( t ) = a5 ( y5 ( t ) − y1 ( t ) ) , y1 ′ ( 0 ) = 0 , y5 ′ ( 0 ) = 0 , y1 ( 0 ) = 0. , y5 ( 0 ) = 12 } , { y1 , y5 } , { t , 0. , 72500. } \left.\text{y1}''(t)=\text{a1}(\text{y5}(t)-\text{y1}(t)),\text{y5}''(t)=\text{a5}(\text{y5}(t)-\text{y1}(t)),\text{y1}'(0)=0,\text{y5}'(0)=0,\text{y1}(0)=0.,\text{y5}(0)=12\right\},\{\text{y1},\text{y5}\},\{t,0.,72500.\}

Then solve for a distance of 3: y5 ( t ) − y1 ( t ) = 3 ⇒ t = 68756.15041989712 \text{y5}(t)-\text{y1}(t)=3 \Rightarrow t=68756.15041989712 .

Then, substitute that time into the solution of the differential equations to get the distance traveled: the lighter mass moves 7.5 R and the heavier mass moves 1.5 R. Note, the distances traveled ratios is the inverse of the mass ratios., which I suspected but did not know for certain.

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