Worked examples · Sheet 1

Gravitational forces between point-like and spherical bodies

Theory used: Law of universal gravitation Superposition principle Field of a homogeneous sphere

Exercise 1

Two homogeneous lead spheres, of masses $m_{1} = 5.0 kg$ and $m_{2} = 8.0 kg$ , have their centers at a distance $r = 0.20 m$ . Compute the magnitude of the gravitational force between the spheres, compare it with the weight of the first sphere and compute the acceleration that the force would give the first sphere if it were free to move. Use $G = 6.67 \cdot 1 0^{- 11} N \cdot m^{2} /k g^{2}$ .

F = G \frac{m _{1} m _{2}}{r ^{2}}, P_{1} = m_{1} g, a_{1} = \frac{F}{m _{1}} = G \frac{m _{2}}{r ^{2}}

F = 6.67 \cdot 1 0^{- 11} \cdot \frac{5.0 \cdot 8.0}{0.2 0 ^{2}} N = 6.67 \cdot 1 0^{- 11} \cdot 1000 N = 6.7 \cdot 1 0^{- 8} N

P_{1} = 5.0 \cdot 9.81 N = 49 N, \frac{F}{P _{1}} = \frac{6.67 \cdot 1 0 ^{- 8}}{49.05} = 1.4 \cdot 1 0^{- 9}

a_{1} = \frac{6.67 \cdot 1 0 ^{- 8} N}{5.0 kg} = 1.3 \cdot 1 0^{- 8} m/ s^{2}

The spheres are homogeneous, so by the field of a homogeneous sphere each one attracts external points as if its mass were concentrated at the center; for this reason the law of universal gravitation uses the distance between the centers, even though the spheres are not small compared with $0.20 m$ . The force is about $6.7 \cdot 1 0^{- 8} N$ , a little more than a billionth of the weight of the sphere. The acceleration is obtained from Newton’s second law by dividing the force by $m_{1}$ , and the mass $m_{1}$ cancels: the acceleration of a body depends only on the mass that attracts it.

As a dimensional check, $N \cdot m^{2} /k g^{2} \cdot k g^{2} / m^{2} = N$ . The result explains why the attraction between everyday objects is not observed: a torsion balance, like Cavendish’s, is needed to measure it.

Exercise 2

Three point masses lie on an $x$ axis: $m_{1} = 4.0 kg$ at $x_{1} = 0$ , $m_{2} = 1.0 kg$ at $x_{2} = 0.50 m$ and $m_{3} = 9.0 kg$ at $x_{3} = 1.50 m$ . Compute the total gravitational force on $m_{2}$ .

F_{12} = G \frac{m _{1} m _{2}}{( x _{2} - x _{1} ) ^{2}} toward - x, F_{32} = G \frac{m _{3} m _{2}}{( x _{3} - x _{2} ) ^{2}} toward + x

F_{x} = - F_{12} + F_{32} = G m_{2} (\frac{m _{3}}{( x _{3} - x _{2} ) ^{2}} - \frac{m _{1}}{( x _{2} - x _{1} ) ^{2}})

F_{x} = 6.67 \cdot 1 0^{- 11} \cdot 1.0 \cdot (\frac{9.0}{1. 0 ^{2}} - \frac{4.0}{0.5 0 ^{2}}) N = 6.67 \cdot 1 0^{- 11} \cdot (9.0 - 16) N = - 4.7 \cdot 1 0^{- 10} N

By the superposition principle the total force on $m_{2}$ is the sum of the forces due to $m_{1}$ and to $m_{3}$ , each computed as if the other mass were not there. Each force is attractive, so it is directed toward the mass that exerts it: the force of $m_{1}$ , which lies to the left of $m_{2}$ , points toward $- x$ , while that of $m_{3}$ points toward $+ x$ . The two forces have the same line of action, therefore the sum reduces to the sum of the $x$ components with their signs.

The mass $m_{3}$ is larger, but it is at twice the distance and the force decreases with the square of the distance. The contribution of $m_{1}$ prevails, so the total force has magnitude $4.7 \cdot 1 0^{- 10} N$ and is directed toward $m_{1}$ .

Exercise 3

A point mass $m = 0.50 kg$ is at the origin of a frame $O x y$ . A mass $m_{A} = 6.0 kg$ is at the point $A = (0.30 m, 0)$ and a mass $m_{B} = 8.0 kg$ at the point $B = (0, 0.40 m)$ . Compute the magnitude and direction of the total gravitational force on $m$ .

Positions in decimeters; the arrows have length proportional to the magnitude of the forces, with

F_{A} : F_{B} = 4 : 3

F = F_{A} + F_{B}, F_{x} = G \frac{m m _{A}}{x _{A}^{2}}, F_{y} = G \frac{m m _{B}}{y _{B}^{2}}

F = F_{x}^{2} + F_{y}^{2}, tan φ = \frac{F _{y}}{F _{x}}

F_{x} = 6.67 \cdot 1 0^{- 11} \cdot \frac{0.50 \cdot 6.0}{0.3 0 ^{2}} N = 2.22 \cdot 1 0^{- 9} N

F_{y} = 6.67 \cdot 1 0^{- 11} \cdot \frac{0.50 \cdot 8.0}{0.4 0 ^{2}} N = 1.67 \cdot 1 0^{- 9} N

F = 2.2 2^{2} + 1.6 7^{2} \cdot 1 0^{- 9} N = 2.78 \cdot 1 0^{- 9} N, tan φ = \frac{1.67}{2.22} = 0.75, φ = 36. 9^{\circ}

The two masses lie on the axes, so the force of $m_{A}$ is directed along $+ x$ and that of $m_{B}$ along $+ y$ , toward the masses that exert them. By the superposition principle the total force is their vector sum, and since the two forces are perpendicular their magnitudes are directly the components $F_{x}$ and $F_{y}$ of the resultant.

The magnitude is obtained with the Pythagorean theorem and the angle $φ$ with the $x$ axis from the tangent. The numbers are in the ratio $F_{x} : F_{y} = \frac{100}{3} : 25 = 4 : 3$ , so the triangle of forces has sides proportional to $3$ , $4$ , $5$ and $tan φ = \frac{3}{4}$ exactly. The total force has magnitude $2.78 \cdot 1 0^{- 9} N$ and forms an angle of $36. 9^{\circ}$ with the $x$ axis, toward the first quadrant.