Prove that, if theorbital speed of the moon increases by 42% , it would stop orbiting around the earth.

Prove that, if theorbital speed of the moon increases by 42% , it woul

1.	Prove that, if theorbital speed of the moon increases by 42% , it would stop orbiting around the earth.
Answer» Solution :Let mass of the earth =M and radius of the moon.s orbit =r Orbital speed of the moon `v=sqrt((GM)/r)`. The acceleration DUE to GRAVITY, `g^.=(GM)/(r^2) or, GM=g^.r^2` `therefore v=sqrt((g^.r^2)/r)=sqrt(g^.r)` Also the ESCAPE velocity of the moon with respect to the earth.s surface `v_e=sqrt(2g^.r)` `therefore (v_e)/(v)=sqrt((2g^.r)/(g^.r))=sqrt(2)=1.414` This implies, `v_e=1.414v=141.4%` of orbital speed v . Hence , if the orbital speed ofthe moon increases by 42% , which means that it changes to 142% of its present value ,it crosses the valueof the escapevelocity from the earth. As a result, the moon will move out of the earth.s gravitational field and will not move around the earth ANYMORE.