Question

# Two uniform solid spheres held fixed of equal radii R, but mass M and 4M have centre to centre separation 6R. A projectile of mass m is projected from the surface of the sphere of mass M directly towards the centre of the second sphere. Obtain an expression for the minimum speed v of the projectile so that it reaches the surface of the second sphere.

A
(3GM5R)1/2
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B
(5GM3R)1/2
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C
(2GM5R)1/2
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D
(5GM2R)1/2
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Solution

## The correct option is A (3GM5R)1/2 Let N be the null point between sphere, which is at distance x from the centre of M. At null point, field due to mass M will be equal to mass 4M, ∴EM=E4M ⇒GMx2=G×4M(6R−x)2 ⇒1x=26R−x ⇒x=2R (Neglecting −ve value) If m just crosses null point N with zero velocity, it will be pulled towards 4M due to its gravitational attraction. Thus, the minimum speed will be such that m just reaches N with zero speed. Since external and non-conservative force are absent in this scenario, mechanical energy of the system (M+4M+m) will be conserved. Now, conserving mechanical energy between the point when m is at the surface of M and when it is at the null point. Ui+Ki=Uf+Kf ⇒−G(M)(m)R−G(4M)(m)5R+12mv2=−G(M)(m)2R−G(4M)(m)4R+0 ⇒−9GMm5R+12mv2=−3GMm2R ⇒12mv2=95GMmR−3GMm2R=310GMmR ∴v=√3GM5R Hence, option (a) is correct. Why this question ? Key Concept - The particle just has to cross the null point in order to reach the other particle. This is a frequently asked concept.

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