Gravitation and the Principle of Superposition Problems and Solutions

Problem#1

An 8.00-kg point mass and a 15.0-kg point mass are held in place 50.0 cm apart. A particle of mass m is released from a point between the two masses 20.0 cm from the 8.00-kg mass along the line connecting the two fixed masses. Find the magnitude and direction of the acceleration of the particle.

Answer:
Let +x be toward the 15.0 kg mass.
The magnitude of the interaction force between mass 3 and mass 1 and mass 2 is respectively

Fig.9

F₃₁ = Gm₁m₃/r₃₁² = (6.67 x 10⁻¹¹ Nm²/kg²)(8 kg)m₃/(0.200 m)² = 1.334 x 10⁻⁸ m₃N/kg in the (─x) direction

F₃₂ = Gm₂m₃/r₃₂² = (6.67 x 10⁻¹¹ Nm²/kg²)(15 kg)m₃/(0.400 m)² = 1.112 x 10⁻⁸ m₃N/kg in the (+x) direction

The net force is

F_net,x = F₃₁ + F₃₂
F_net,x = ─1.334 x 10⁻⁸ m₃N/kg + 1.112 x 10⁻⁸ m₃N/kg = ─2.2 x 10⁻⁸ m₃N/kg

Using Newton’s second law,

a_x = ∑F_net,x/m₃
a_x = (─2.2 x 10⁻⁸ m₃N/kg)/m₃ = ─2.2 x 10⁻⁸ N/kg

Problem#2
In Fig. 10a, a square of edge length 20.0 cm is formed by four spheres of masses m₁ = 5.00 g, m₂ = 3.00 g, m₃ = 1.00 g, and m₄ = 5.00 g. In unit-vector notation, what is the net gravitational force from them on a central sphere with mass m₅ = 2.50 g?

Fig.10

Answer:

From the superposition principle, we have:

F_net = F₅₁ + F₅₂ + F₅₃ + F₅₄

The gravitational forces │F₅₁│ = │F₅₄│ on m₅ from the two 5.00 g masses m₁ and m₄ cancel each other. Contributions to the net force on m₅ come from the remaining two masses:

F_net =  F₅₂ + F₅₃
F_net =  Gm₅m₂/r₅₂² + [─Gm₅m₃/r₅₃²]
F_net =  [Gm₅/r₅₂²][m₂ ─ m₃]
F_net =  [(6.67 x 10⁻¹¹ Nm²/kg²)(2.50 x 10⁻³ kg)/(√2 x 10⁻¹ m)²][3.00 x 10⁻³ kg ─ 1.00 x 10⁻³ kg]
F_net =  1.67 x 10⁻¹⁴ N

The force is directed along the diagonal between m₂ and m₃, towards m₂. In unit-vector notation, we have

F_net = F_net(cos 45⁰i + cos 45⁰j) = 1.18 x 10⁻¹⁴ N(i + j)

Problem#3
As seen in Fig. 11a, two spheres of mass m and a third sphere of mass M form an equilateral triangle, and a fourth sphere of mass m4 is at the center of the triangle. The net gravitational force on that entral sphere from the three other spheres is zero. (a) What is M in terms of m? (b) If we double the value of m₄, what then is the magnitude of the net gravitational force on the central sphere?

Fig.11

Answer:

(a) The picture above shows all four masses and three gravitational forces acting from each of the masses to mass m₄ . The choice of the coordinate axes is also shown in this picture. The net force acting on m₄ is zero, which means that

∑F_x = 0 and ∑F_y = 0

Since both masses m are in symmetric positions with respect to mass m₄ , they both provide the same components of forces in y-direction, and the same in magnitude but opposite in sign components in x-direction. Moreover, the force acting between m₄ and M has only y-component, so the equation for the x-direction does not involve any unknowns and we do not need it to solve the problem. The only equation necessary is the equation for y-direction. Let r be the distance between the central sphere and any of the other three spheres. It is the same for all of them, since this is the equilateral triangle. This means that the gravitational force between the mass m₄ and any of the masses m is

F_m = Gm₄m/r²

and between mass m₄ and mass M is

F_M = Gm₄M/r²

The equation for the forces' components in y-direction becomes

F_M ─ 2F_my = 0

(Gm₄M/r²) ─ (2Gm₄m/r²) cos 60⁰ = 0

M ─ 2m x ½ = 0
So, M = m

(b) if one doubles mass m₄ it will double all three gravitational forces acting on it. Since they all will change in proportional manner this will neither change their directions, nor the fact that they all are of the same magnitude. This means that the net gravitational force will stay zero in this case

Problem#4
Figure 12 shows a spherical hollow inside a lead sphere of radius R = 4.00 cm; the surface of the  hollow passes through the center of the sphere and “touches” the right side of the sphere. The mass of the sphere before hollowing was M = 2.95 kg.With what gravitational force does the hollowed-out lead sphere attract a small sphere of mass m = 0.431 kg that lies at a distance d = 9.00 cm from the center of the lead sphere, on the straight line connecting the centers of the spheres and of the hollow?

Fig.12

Answer:

This is a job for negative mass. We take the force the solid sphere would exert, then subtract the force of the material which makes up the hollow. In effect, we treat the hollow as negative mass.

F = GMm/d² – GM_hm/(d − R/2)²

Since the radius of the hollow region is half of that of the sphere, we obtain for M_h:

M_h =ρ⋅4π/3⋅(R/2)³= M/8

Finally, we obtain:

F = GMm/d² – G(M/8)m/(d − R/2)²
F = GMm[1/d²− ½(2d − R)²]
F = (6.67 x 10⁻¹¹ Nm²/kg²)(2.95 kg)(0.431 kg)[1/(0.09 m)²− ½(2 x 0.09 m – 0.04 m)²]
F = 8.31 × 10⁻⁹ N

Problem#5
A solid sphere of mass m and radius r is placed inside a hollow thin spherical shell of mass M and radius R as shown in figure (13). A particle of mass m' is placed on the line joining the two centers at a distance x from the point of contact of the sphere and the shell. Find the magnitude of the resultant gravitational force on this particle due to the sphere and the shell if (a) r < x < 2r, (b) 2r < x < 2R and
(c) x > 2R.

Fig.13

Answer:

For situations in (a) and (b) x < 2R, the shell does not force the particle m'

(a) if r < x < 2r (Fig.14a)

The distance of the particle m' from the center of the solid sphere d = x - r
The resultant gravitational force on m' will be the force due to the solid sphere of radius s = x – r

F = Gmm'd/r³
F = Gmm'(x – r)/r³

(b) If 2r < x < 2R (Fig.14b)
In this case, the particle m' is between the space of the solid sphere and the shell. The resultant gravitational force on the particle will be due to the solid sphere only and its magnitude will be

F = Gmm'/d², here d = x - r, the distance of the particle from the center of the solid sphere

F = Gmm'/(x – r)²

(c) If x > 2R, (Fig.14c)

Fig.14

Now both the shell and the solid sphere will exert the gravitational force on the particle m'. The masses of the sphere and the shell will be assumed to be concentrated at their respective centers.

Force by the solid sphere:
The distance of the particle from the center of the solid sphere is d  = x - r
Hence force

F = Gmm'/s² = Gmm'/(x – r)²

Force by the shell:
The distance of the particle from the center of the shell

s = x – R
Hence force
F = GMm'/s²
F = GMm'/(x – R)²

Since both of these forces are in the same direction (towards the line joining two centers) hence the magnitude of the resultant force will be added, i.e.
The resultant gravitational force on m'

F = Gmm'/(x – r)² + GMm'/(x – R)²
F = Gm’[m/(x – r)² + GM/(x – R)²] 

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