Solved Examples on Gravitation & Projectile

Question 1:-At what initial speed must the basketball player throw the ball, at 55º above the horizontal, to make the foul shot, as shown in below figure? The basketball rim is 18 in. in diameter. Obtain other data from below figure

Concept:The figure below shows the basketball player making a foul shot:

It can be seen from the figure that the effective horizontal distance to make that shot is

14ft -1ft=13ft whereas the effective vertical distance travelled by the ball is 10 ft - 7 ft = 3 ft

Assume that the angle at which the ball is released with initial velocity

v_{0}isθsuch that the horizontal component of initial velocity isv_{0x}=v_{0 }cosθand the vertical component of initial velocity isv_{0y}=v_{0 }sinθ.The time taken by the ball to reach the basket can be written in terms of horizontal component of velocity and the effective horizontal distance travelled by the ball as:

t= 13 ft /v_{0x}Substitute

v_{0x}=v_{0 }cosθ,

t= 13 ft /v_{0 }cosθ…… (1)Similarly, one can write the time

tin terms of the vertical component on initial velocity and the effective vertical distance travelled by the ball as:3ft =

v_{0y}t– ½gt^{2}Substitute

v_{0y}=v_{0}sinθ,3ft = (

v_{0}sinθ)t– ½gt^{2}Substitute

t= 13 ft/v_{0 }cosθfrom equation (1),

Substitute the value ofθandgto obtain the value ofv_{0}.

Solution:Substitute 55º for

θand 32.2 ft/s^{2}forgin equation

Therefore, the magnitude of initial velocity of the ball is 23 ft/s.

Question 2:-A spherical hollow is made in a lead sphere of radiusR, such that its surface touches the outside surface of the lead sphere and passes through its center (see below figure). The mass of the sphere before hollowing wasM. With what force, according to the law of universal gravitation, will the hollowed lead sphere attract a small sphere of massm, which lies at a distancedfrom the center of the lead sphere on the straight line connecting the centers of the spheres and of the follow? [1946, Olympic, Moscow State University]

Concept:The force of gravity on the small sphere of mass due to the lead sphere after it is made hollow is equal to the force of gravity from a solid lead sphere minus the force which would have been contributed by the smaller lead sphere which would have filled the hole.

Solution:From the Newton’s law of universal gravitation, the force of attraction between the lead sphere and the small sphere is

F_{1}=GMm/d^{2}Here, gravitational constant is

G, total mass of the lead sphere isM, mass of the small sphere ismand distance between the centers of the spheres isd.So, we need to know about the size and mass of the lead which was removed to make the hole.

The density of the lead is given by

ρ=M/[(4/3)(πR^{3})]Here, radius of the lead sphere is

R.The spherical sphere has a radius of

R/2.If the density is constant, then the mass of the hole will be the density times the volume of the hollow sphere. Thus, the mass of the hollow sphere is

The hole is closer to the small sphere. This means the center of the hole is

d-(R/2) away from the small sphere.Now, the gravitational force between the hollow sphere and the small sphere is

The force of attraction with which the lead sphere after it is made hollow and the small sphere is equal to the force of the whole lead sphere before it is made hollow minus the force of the hollow lead sphere. The corresponding force is

Therefore, the required force from the lead sphere which attract the small sphere is

Question 3:-A certain triple-star system consists of two stars, each of massm, revolving about a central star; massM, in the same circular orbit. The two stars stay at opposite ends of a diameter of the circular orbit; see below figure. Derive an expression for the period of revolution of the stars; the radius of the orbit isr.

Concept:The gravitational force due to the two stars to the central star is

F_{1}=GMm/r^{2}+GMm/r^{2}= 2

GMm/r^{2}Here, gravitational constant is

G, mass of the central star isM, mass of the orbiting stars ismand radius of the orbit isr.Also, the gravitational force between the orbiting stars is,

F_{2}=Gmm/(2r)^{2}

^{ }=Gm^{2}/4r^{2}The net gravitational force acting on the stars is equal to the sum of the gravitational forces acting on the system. It is given as,

F

_{net}= F_{1}+F_{2}= 2

GMm/r^{2}+Gm^{2}/4r^{2}=

Gm/r^{2}[(2M)+m/4]The centripetal force that makes the stars to orbit around the central star is

F_{c}= 4π^{2}mr/T^{2}Here, period of the stars around the central star is

T.

Solution:The centripetal force is provided by the net gravitational force acting on the system of stars. Thus,

F_{net}=F_{c}Substitute

Gm/r^{2}[(2M)+m/4] forF_{net}and 4π^{2}mr/T^{2}forF_{c}in the above equation.

F_{net}=F_{c}

Gm/r^{2}[(2M)+m/4] = 4π^{2}mr/T^{2}So,

T= √16π^{2}r^{3}/[G(8M+m)]Therefore, the expression for the period of revolution of the stars is √16π

^{2}r^{3}/[G(8M+m)].