Gravitational Force and Potential Energy: A Comprehensive Guide with Examples

Posted on Aug 27, 2024 in Physics

4. The gravitational force between you and the moon at its initial position (directly opposite of Earth from you) is

where

 is the mass of the moon,

 is the distance between the moon and the Earth, and

 is the radius of the Earth. At its final position (directly above you), the gravitational force between you and the moon is

.

(a) The ratio of the moon’s gravitational pulls at the two different positions is

Therefore, the increase is 0.06898, or approximately, 6.9%.

(b) The change of the gravitational pull may be approximated as

On the other hand, your weight, as measured on a scale on Earth is

.

Since the moon pulls you “up,” the percentage decrease of weight is

5. We require the magnitude of force (given by Eq. 13-1) exerted by particle C on A be equal to that exerted by B on A.  Thus,

 =  .

We substitute in m_B = 3m_A   and m_B = 3m_A, and (after canceling “m_A”) solve for r. We find r = 5d.  Thus, particle C is placed on the x axis, to left of particle A (so it is at a negative value of x), at x = –5.00d.

9. The gravitational force from Earth on you (with mass m) is

where

 If r is the distance between you and a tiny black hole of mass

 that has the same gravitational pull on you as the Earth, then

Combining the two equations, we obtain

11. (a) The distance between any of the spheres at the corners and the sphere at the center is

where

 is the length of one side of the equilateral triangle. The net (downward) contribution caused by the two bottom-most spheres (each of mass m) to the total force on m₄ has magnitude

This must equal the magnitude of the pull from M, so

which readily yields m = M.

(b) Since m₄ cancels in that last step, then the amount of mass in the center sphere is not relevant to the problem. The net force is still zero.

13. If the lead sphere were not hollowed the magnitude of the force it exerts on m would be F₁ = GMm/d². Part of this force is due to material that is removed. We calculate the force exerted on m by a sphere that just fills the cavity, at the position of the cavity, and subtract it from the force of the solid sphere.

The cavity has a radius r = R/2. The material that fills it has the same density (mass to volume ratio) as the solid sphere. That is M_c/r³= M/R³, where M_c is the mass that fills the cavity. The common factor 4p/3 has been canceled. Thus,

The center of the cavity is d – r = d – R/2 from m, so the force it exerts on m is

The force of the hollowed sphere on m is

16. Since the rod is an extended object, we cannot apply Equation 13-1 directly to find the force. Instead, we consider a small differential element of the rod, of mass

 of thickness

 at a distance r from

. The gravitational force between

 and

is

,

where we have substituted

 since mass is uniformly distributed. The direction of

 is to the right (see figure). The total force can be found by integrating over the entire length of the rod:

.

Substituting the values given in the problem statement, we obtain

21. (a) The gravitational acceleration is

(b) Note that the total mass is 5M. Thus,

23. From Eq. 13-14, we see the extreme case is when “g” becomes zero, and plugging in Eq. 13-15 leads to

Thus, with R = 20000 m and w = 2p rad/s, we find M = 4.7 ´ 10²⁴ kg » 5 ´ 10²⁴ kg.

25. (a) The magnitude of the force on a particle with mass m at the surface of Earth is given by F = GMm/R², where M is the total mass of Earth and R is Earth’s radius. The acceleration due to gravity is

(b) Now a_g = GM/R², where M is the total mass contained in the core and mantle together and R is the outer radius of the mantle (6.345 ´ 10⁶ m, according to Fig. 13-43). The total mass is

M = (1.93 ´ 10²⁴ kg + 4.01 ´ 10²⁴ kg ) = 5.94 ´ 10²⁴ kg.

The first term is the mass of the core and the second is the mass of the mantle. Thus,

(c) A point 25 km below the surface is at the mantle-crust interface and is on the surface of a sphere with a radius of R = 6.345 ´ 10⁶ m. Since the mass is now assumed to be uniformly distributed the mass within this sphere can be found by multiplying the mass per unit volume by the volume of the sphere:

 where M_e is the total mass of Earth and R_e is the radius of Earth. Thus,

The acceleration due to gravity is

32. (a) The potential energy at the surface is (according to the graph) –5.0 ´ 10⁹ J, so (since U is inversely proportional to r – see Eq. 13-21) at an r-value a factor of 5/4 times what it was at the surface then U must be a factor of 4/5 what it was.  Thus, at r = 1.25R_s U = – 4.0 ´ 10⁹ J.  Since mechanical energy is assumed to be conserved in this problem, we have K + U = –2.0 ´ 10⁹ J at this point.  Since U = – 4.0 ´ 10⁹ J here, then

 at this point.

(b) To reach the point where the mechanical energy equals the potential energy (that is, where U = – 2.0 ´ 10⁹ J) means that U must reduce (from its value at r = 1.25R_s) by a factor of 2 – which means the r value must increase (relative to r = 1.25R_s) by a corresponding factor of 2.  Thus, the turning point must be at r = 2.5R_s .

35. (a) The work done by you in moving the sphere of mass m_B equals the change in the potential energy of the three-sphere system. The initial potential energy is

and the final potential energy is

The work done is

(b) The work done by the force of gravity is -(U_f – U_i) = -5.0 ´ 10^-13 J.

37. (a) We use the principle of conservation of energy. Initially the particle is at the surface of the asteroid and has potential energy U_i = –GMm/R, where M is the mass of the asteroid, R is its radius, and m is the mass of the particle being fired upward. The initial kinetic energy is

. The particle just escapes if its kinetic energy is zero when it is infinitely far from the asteroid. The final potential and kinetic energies are both zero. Conservation of energy yields

–GMm/R + ½mv² = 0.

We replace GM/R with a_gR, where a_g is the acceleration due to gravity at the surface. Then, the energy equation becomes –a_gR + ½v² = 0. We solve for v:

(b) Initially the particle is at the surface; the potential energy is U_i = –GMm/R and the kinetic energy is K_i = ½mv². Suppose the particle is a distance h above the surface when it momentarily comes to rest. The final potential energy is U_f = –GMm/(R + h) and the final kinetic energy is K_f = 0. Conservation of energy yields

We replace GM with a_gR² and cancel m in the energy equation to obtain

The solution for h is

(c) Initially the particle is a distance h above the surface and is at rest. Its potential energy is U_i = –GMm/(R + h) and its initial kinetic energy is K_i = 0. Just before it hits the asteroid its potential energy is U_f = –GMm/R. Write

 for the final kinetic energy. Conservation of energy yields

We substitute a_gR² for GM and cancel m, obtaining

The solution for v is

39. (a) The momentum of the two-star system is conserved, and since the stars have the same mass, their speeds and kinetic energies are the same. We use the principle of conservation of energy. The initial potential energy is U_i = –GM²/r_i, where M is the mass of either star and r_i is their initial center-to-center separation. The initial kinetic energy is zero since the stars are at rest. The final potential energy is U_f = -2GM²/r_i since the final separation is r_i/2. We write Mv² for the final kinetic energy of the system. This is the sum of two terms, each of which is ½Mv². Conservation of energy yields

The solution for v is

(b) Now the final separation of the centers is r_f = 2R = 2 ´ 10⁵ m, where R is the radius of either of the stars. The final potential energy is given by U_f = –GM²/r_f and the energy equation becomes –GM²/r_i = –GM²/r_f + Mv². The solution for v is

42. (a) Applying Eq. 13-21 and the Pythagorean theorem leads to

       U =  –

where M is the mass of particle B (also that of particle C) and m is the mass of particle A.  The value given in the problem statement (for infinitely large y, for which the second term above vanishes) determines M, since D is given.  Thus M = 0.50 kg.

(b) We estimate (from the graph) the y = 0 value to be U_o = – 3.5 × 10^-10 J.  Using this, our expression above determines m.  We obtain m =1.5 kg.

45. Let N be the number of stars in the galaxy, M be the mass of the Sun, and r be the radius of the galaxy. The total mass in the galaxy is N M and the magnitude of the gravitational force acting on the Sun is F = GNM²/r². The force points toward the galactic center. The magnitude of the Sun’s acceleration is a = v²/R, where v is its speed. If T is the period of the Sun’s motion around the galactic center then v = 2pR/T and a = 4p²R/T². Newton’s second law yields GNM²/R² = 4p²MR/T². The solution for N is

The period is 2.5 ´ 10⁸ y, which is 7.88 ´ 10¹⁵ s, so

49. (a) If r is the radius of the orbit then the magnitude of the gravitational force acting on the satellite is given by GMm/r², where M is the mass of Earth and m is the mass of the satellite. The magnitude of the acceleration of the satellite is given by v²/r, where v is its speed. Newton’s second law yields GMm/r² = mv²/r. Since the radius of Earth is 6.37 ´ 10⁶ m the orbit radius is r = (6.37 ´ 10⁶ m + 160 ´ 10³ m) = 6.53 ´ 10⁶ m. The solution for v is

(b) Since the circumference of the circular orbit is 2pr, the period is

This is equivalent to 87.5 min.

53. (a) If we take the logarithm of Kepler’s law of periods, we obtain

where we are ignoring an important subtlety about units (the arguments of logarithms cannot have units, since they are transcendental functions). Although the problem can be continued in this way, we prefer to set it up without units, which requires taking a ratio. If we divide Kepler’s law (applied to the Jupiter-moon system, where M is mass of Jupiter) by the law applied to Earth orbiting the Sun (of mass M_o), we obtain

where T_E = 365.25 days is Earth’s orbital period and r_E = 1.50 ´ 10¹¹ m is its mean distance from the Sun. In this case, it is perfectly legitimate to take logarithms and obtain

(written to make each term positive) which is the way we plot the data (log (r_E/a) on the vertical axis and log (T_E/T) on the horizontal axis).

(b) When we perform a least-squares fit to the data, we obtain

log (r_E/a) = 0.666 log (T_E/T) + 1.01,

which confirms the expectation of slope = 2/3 based on the above equation.

(c) And the 1.01 intercept corresponds to the term 1/3 log (M_o/M) which implies

Plugging in M_o = 1.99 ´ 10³⁰ kg (see Appendix C), we obtain M = 1.86 ´ 10²⁷ kg for Jupiter’s mass. This is reasonably consistent with the value 1.90 ´ 10²⁷ kg found in Appendix C.

54. (a) The period is T = 27(3600) = 97200 s, and we are asked to assume that the orbit is circular (of radius r = 100000 m). Kepler’s law of periods provides us with an approximation to the asteroid’s mass:

(b) Dividing the mass M by the given volume yields an average density equal to

r = 6.3 ´ 10¹⁶/1.41 ´ 10¹³ = 4.4 ´ 10³ kg/m³,

which is about 20% less dense than Earth.

54. (a) The period is T = 27(3600) = 97200 s, and we are asked to assume that the orbit is circular (of radius r = 100000 m). Kepler’s law of periods provides us with an approximation to the asteroid’s mass:

(b) Dividing the mass M by the given volume yields an average density equal to

r = 6.3 ´ 10¹⁶/1.41 ´ 10¹³ = 4.4 ´ 10³ kg/m³,

which is about 20% less dense than Earth.

55. In our system, we have m₁ = m₂ = M (the mass of our Sun, 1.99 ´ 10³⁰ kg). With r = 2r₁ in this system (so r₁ is one-half the Earth-to-Sun distance r), and v = pr/T for the speed, we have

With r = 1.5 ´ 10¹¹ m, we obtain T = 2.2 ´ 10⁷ s. We can express this in terms of Earth-years, by setting up a ratio:

55. In our system, we have m₁ = m₂ = M (the mass of our Sun, 1.99 ´ 10³⁰ kg). With r = 2r₁ in this system (so r₁ is one-half the Earth-to-Sun distance r), and v = pr/T for the speed, we have

With r = 1.5 ´ 10¹¹ m, we obtain T = 2.2 ´ 10⁷ s. We can express this in terms of Earth-years, by setting up a ratio:

59. Each star is attracted toward each of the other two by a force of magnitude GM²/L², along the line that joins the stars. The net force on each star has magnitude 2(GM²/L²) cos 30° and is directed toward the center of the triangle. This is a centripetal force and keeps the stars on the same circular orbit if their speeds are appropriate. If R is the radius of the orbit, Newton’s second law yields (GM²/L²) cos 30° = Mv²/R.

The stars rotate about their center of mass (marked by a circled dot on the diagram above) at the intersection of the perpendicular bisectors of the triangle sides, and the radius of the orbit is the distance from a star to the center of mass of the three-star system. We take the coordinate system to be as shown in the diagram, with its origin at the left-most star. The altitude of an equilateral triangle is

, so the stars are located at x = 0, y = 0; x = L, y = 0; and x = L/2,

. The x coordinate of the center of mass is x_c = (L + L/2)/3 = L/2 and the y coordinate is

. The distance from a star to the center of mass is

.

Once the substitution for R is made Newton’s second law becomes

. This can be simplified somewhat by recognizing that

, and we divide the equation by M. Then, GM/L² = v²/L and

.

67. (a) The force acting on the satellite has magnitude GMm/r², where M is the mass of Earth, m is the mass of the satellite, and r is the radius of the orbit. The force points toward the center of the orbit. Since the acceleration of the satellite is v²/r, where v is its speed, Newton’s second law yields GMm/r² = mv²/r and the speed is given by v =

. The radius of the orbit is the sum of Earth’s radius and the altitude of the satellite: r = (6.37 ´ 10⁶ + 640 ´ 10³) m = 7.01 ´ 10⁶ m. Thus,

(b) The period is

T = 2pr/v = 2p(7.01 ´ 10⁶ m)/(7.54 ´ 10³ m/s) = 5.84 ´ 10³ s

 97 min.

(c) If E₀ is the initial energy then the energy after n orbits is E = E₀ – nC, where C = 1.4 ´ 10⁵ J/orbit. For a circular orbit the energy and orbit radius are related by E = –GMm/2r, so the radius after n orbits is given by r = –GMm/2E.

The initial energy is

the energy after 1500 orbits is

and the orbit radius after 1500 orbits is

The altitude is h = r – R = (6.78 ´ 10⁶ m – 6.37 ´ 10⁶ m) = 4.1 ´ 10⁵ m. Here R is the radius of Earth. This torque is internal to the satellite-Earth system, so the angular momentum of that system is conserved.

(d) The speed is

(e) The period is

93 min.

(f) Let F be the magnitude of the average force and s be the distance traveled by the satellite. Then, the work done by the force is W = –Fs. This is the change in energy: –Fs = DE. Thus, F = -DE/s. We evaluate this expression for the first orbit. For a complete orbit s = 2pr = 2p(7.01 ´ 10⁶ m) = 4.40 ´ 10⁷ m, and DE = -1.4 ´ 10⁵ J. Thus,

(g) The resistive force exerts a torque on the satellite, so its angular momentum is not conserved.

(h) The satellite-Earth system is essentially isolated, so its momentum is very nearly conserved.