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Worked Examples · Example 7.4

Q.Two uniform solid spheres of equal radii RR, but mass MM and 4M4M have a centre to centre separation 6R6R, as shown in Fig. 7.10. The two spheres are held fixed. A projectile of mass mm is projected from the surface of the sphere of mass MM directly towards the centre of the second sphere. Obtain an expression for the minimum speed vv of the projectile so that it reaches the surface of the second sphere.

Figure 7.10
Figure 7.10
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The projectile needs just enough speed to crest the gravitational potential 'hill' between the two spheres — the point where the pulls from the two spheres exactly balance — after which the second sphere's gravity carries it the rest of the way. Using conservation of energy between the launch point and that balance point gives a minimum speed of v=3GM5Rv=\sqrt{\dfrac{3GM}{5R}}.

Setting up

Place the centre of the sphere of mass MM at x=0x=0 and the centre of the sphere of mass 4M4M at x=6Rx=6R; both spheres have radius RR, so their surfaces are at x=Rx=R and x=5Rx=5R. The projectile is launched from x=Rx=R (the surface of the MM-sphere) toward the other sphere.

Finding the neutral point

Between the spheres, the projectile is pulled left by MM and right by 4M4M. These pulls balance at the point where

GMmx2=G(4M)m(6R−x)2\frac{GMm}{x^2} = \frac{G(4M)m}{(6R-x)^2}

1x2=4(6R−x)2  ⇒  1x=26R−x  ⇒  6R−x=2x  ⇒  x=2R\frac{1}{x^2} = \frac{4}{(6R-x)^2} \;\Rightarrow\; \frac{1}{x} = \frac{2}{6R-x} \;\Rightarrow\; 6R-x = 2x \;\Rightarrow\; x=2R

This neutral point NN (at x=2Rx=2R) is where the projectile's gravitational potential energy is at its highest along the path — beyond it, the pull from the 4M4M sphere dominates and pulls the projectile the rest of the way in.

Note

The minimum-speed condition is that the projectile just reaches this neutral point with zero leftover speed — not that it reaches the far sphere's surface with zero speed. Once past the neutral point, the stronger sphere's gravity takes over and accelerates it the rest of the way, so no extra launch speed is needed for that part of the journey.

Applying conservation of energy

The potential energy of the projectile at position xx is

U(x)=−GMmx−G(4M)m6R−xU(x) = -\frac{GMm}{x} - \frac{G(4M)m}{6R-x} …

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