But the ans is √27GM÷R. They gave with this velocity the object of mass m strikes heavier mass 4M from neutral point

Asked by venkat 12th December 2017, 2:29 PM

Answered by Expert

Answer:

Refer the figure you sent in your previous question. It is clearly mentioned in text and figure, the required thing is the minimum projection velocity from the surface of first sphere whos mass is M. As per this requirement I worked out in my earlier answer and checked with my colleague also. I hope you understand the solving procedure of this problem and you also check the mathematics involved(integration, algebra etc.)

It is also possible to find the velocity of the object of mass m that strikes heavier mass 4M from neutral point as given below

begin mathsize 12px style 1 half m v squared space equals space integral subscript 0 superscript 3 R end superscript open parentheses fraction numerator 4 G m M over denominator begin display style open parentheses 4 R minus h close parentheses squared end style end fraction space minus space fraction numerator G m M over denominator begin display style open parentheses 2 R plus h close parentheses squared end style end fraction close parentheses d h space equals space G m M open vertical bar fraction numerator begin display style 4 end style over denominator begin display style 4 R minus h end style end fraction minus fraction numerator begin display style open parentheses negative 1 close parentheses end style over denominator begin display style open parentheses 2 R plus h close parentheses end style end fraction close vertical bar subscript 0 superscript 3 R end superscript space space equals space fraction numerator 27 G m M over denominator 10 R end fraction H e n c e space v space equals space square root of fraction numerator 5.4 G M over denominator R end fraction end root end style

In the calculation given above it is assumed that a mass of m travels from neutral point to surface of sphere of mass 4M. Starting velocity is assumed as zero. First term in the integration is due to the attractive force of second sphere of mass 4M. Second term is due to the force of first sphere of mass M. From neutral point to the surface of second sphere, the distance is 3R so the limits of integration.

again you check the mathematics(integration, algebra)

Answered by Expert 13th December 2017, 12:50 PM

Rate this answer