two point masses M and 3M are placed at a distance l apart another point mass M is placed in between the line joining them so that the net gravitational force acting on it due to the masses M and 3M is zero the magnitude of gravitational force acting on I'm on the mass M will be
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According to Newton’s Law of Universal Gravitation, the force between two masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
F=(GmM)/L^2
Force on ‘m’ is zero. as (GmM)/L’^2=(3GmM)/L”^2
As the point mass ‘m’ is between ‘M’ and ‘3M’, L’+L”=L
If we find the solution for L’ and calculate the force between the point masses ‘m’ and ‘M’.
After solving the equations however, I didn’t get the answer as in the options. As the answer is coming to be GMm(4+2√3)/L^2.
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Muthusamy P, M.Sc. Retired as Head of Department of Physics. Government of Tamilnadu, lndia.
Answered Sep 17
let GMm = K , a be the distance between M and m .
K/ a^2 = 3K / ( L- a) ^2.
√3a = ( L- a)
a (1 + √3) = L
a = L / ( 1+ √3)
K/ a^2 = K ( 1+ √3)^2/ L^2
Answer is (a) with a correction .
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Debopriyo Paul, B Tech in Electrical Engineering, National Institute of Technology, Tiruchirappalli, Tamil Nadu, India
Updated Oct 16
Let the mass m be nearer to mass M, than mass 3M as the net force acting on it due to the two masses is 0.As 3M>M,so the point mass m mustn't not be experiencing much force of attraction of 3M.Let the distance between m and M be x.So distance between 3M and m is (l-x ) m.(as total distance between them is l.l.)
Force of attraction between m and M :
F= GMmx2...(i)GMmx2...(i)
Force of attraction between 3M and m :
3GMm(l−x)2...(ii)3GMm(l−x)2...(ii)
Eqns (i) and (ii) are equal,as the net force is zero.So,
GMmx2=3GMm(l−x)2GMmx2=3GMm(l−x)2
i.e, (l−x)2=3x2(l−x)2=3x2
(l−x)=3–√xl−x)=3x (Taking square root )
or
l=3–√x+xl=3x+x
Or
x=l3–√+1l3+1
As x is distance between m and M, you can find the force of attraction between m and M.You will find the answer corresponds to (a)
F=(GmM)/L^2
Force on ‘m’ is zero. as (GmM)/L’^2=(3GmM)/L”^2
As the point mass ‘m’ is between ‘M’ and ‘3M’, L’+L”=L
If we find the solution for L’ and calculate the force between the point masses ‘m’ and ‘M’.
After solving the equations however, I didn’t get the answer as in the options. As the answer is coming to be GMm(4+2√3)/L^2.
374 Views · View Upvoters · Answer requested by Atharva Kale
Upvote· 12
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Comment...
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Muthusamy P, M.Sc. Retired as Head of Department of Physics. Government of Tamilnadu, lndia.
Answered Sep 17
let GMm = K , a be the distance between M and m .
K/ a^2 = 3K / ( L- a) ^2.
√3a = ( L- a)
a (1 + √3) = L
a = L / ( 1+ √3)
K/ a^2 = K ( 1+ √3)^2/ L^2
Answer is (a) with a correction .
212 Views · View Upvoters · Answer requested by Rahul Pandey
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Debopriyo Paul, B Tech in Electrical Engineering, National Institute of Technology, Tiruchirappalli, Tamil Nadu, India
Updated Oct 16
Let the mass m be nearer to mass M, than mass 3M as the net force acting on it due to the two masses is 0.As 3M>M,so the point mass m mustn't not be experiencing much force of attraction of 3M.Let the distance between m and M be x.So distance between 3M and m is (l-x ) m.(as total distance between them is l.l.)
Force of attraction between m and M :
F= GMmx2...(i)GMmx2...(i)
Force of attraction between 3M and m :
3GMm(l−x)2...(ii)3GMm(l−x)2...(ii)
Eqns (i) and (ii) are equal,as the net force is zero.So,
GMmx2=3GMm(l−x)2GMmx2=3GMm(l−x)2
i.e, (l−x)2=3x2(l−x)2=3x2
(l−x)=3–√xl−x)=3x (Taking square root )
or
l=3–√x+xl=3x+x
Or
x=l3–√+1l3+1
As x is distance between m and M, you can find the force of attraction between m and M.You will find the answer corresponds to (a)
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