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Distance between 2 stars is 10a. The masses of these stars are M and 16 M and their radii a and 2a. A body of mass m is fired straight from larger star towards smaller star. What should be its minimum initial speed to reach the surface of smaller star ?

Mansi , 8 Years ago
Grade 12
anser 2 Answers
Vikas TU

Last Activity: 8 Years ago

The object needs enough energy to make it to the position where the gravitational forces from the two stars are equal. Beyond that, it will accelerate towards the smaller star. 

If the distance from the larger star's center is x, then this position is: 
GMm/(10a-x)^2 = G(16M)m/x^2 
x = 8a 

At its starting position, the gravitational potential energies with respect to the two stars are -GMm/(8a) and -G(16M)m/(2a). 
At the balanced position, the gravitational potential energies with respect to the two stars are -GMm/(2a) and -G(16M)m/(8a). 
Moving the object has increased its total potential energy by (45/8)GMm/a. 

Therefore it needs to start with kinetic energy of at least (45/8)GMm/a. 
mv^2/2 = (45/8)GMm/a 
v = 3/2 sqrt(5GM/a) m/s

ankit singh

Last Activity: 4 Years ago

Let P be the point on the line joining the centres of the two planets such that the net field at that point is zero.
Then, r2GM(10ar)2G16M=0
(10ar)2=16r2
10ar=4r
r=2a
Potential at point P,
vp=rGM(10ar)G16M=2aGMa2GM
=2a5GM
Now if the particle projected from the larger planet has enough energy to cross this point, it will reach the smaller planet.
For this, the KE imparted to the body must be just enough to raise its total mechanical energy to a value which is equal to PE at point P, i.e.,
21mv22aG(16M)m8aGMm=mvp
or 2v2a8GM8aGM=2a5GMm
or v2=4a45GM
or vmin=23a5GM.

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