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Chapter 13: Problem 80

Tidal Forces near a Black Hole. An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of \(120 \mathrm{~km}\) from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of \(15.0 \mathrm{~km} .\) The astronaut is positioned inside the spaceship such that one of her \(0.030 \mathrm{~kg}\) ears is \(6.0 \mathrm{~cm}\) farther from the black hole than the center of mass of the spacecraft and the other ear is \(6.0 \mathrm{~cm}\) closer. (a) What is the tension between her ears? Would the astronaut find it difficult to keep from being torn apart by the gravitational forces? (Since her whole body orbits with the same angular velocity, one ear is moving too slowly for the radius of its orbit and the other is moving too fast. Hence her head must exert forces on her cars to keep them in their orbits.) (b) Is the center of gravity of her head at the same point as the center of mass? Explain.

Short Answer

Expert verified
The calculated value from step 2 will give the tidal force or the tension between astronaut's ears. And, the center of gravity of her head is essentially at the same point as the center of mass.

Step by step solution

01

Calculate Gravitational Force at Different Locations

\The gravitational force \(F\) experienced by an object of mass \(m\) at a distance \(r\) from a massive object of mass \(M\) can be given by the formula:\[ F = \frac{G M m}{r^2}\]where \(G\) is the Gravitational constant. Here, we calculate the gravitational force at three positions: at the astronaut, at the closer ear, and at the farther ear. The differences in gravitational forces at these locations will give the tidal forces.
02

Determine the Tidal Forces

\The tidal force \(F_t\) between the ears would be the difference of the gravitational forces at the locations of farther and closer ear, i.e., \[ F_t = \Delta F = F_{far-cm} - F_{cm-close} \]Calculate these values by substituting the respective distances and mass of the ears (0.030 kg) in the gravitational force formula.
03

Answer Part (b)

\The center of mass of her head is the average position of all the mass that makes up her head. The center of gravity of her head, in this context, would also be the same point because the gravitational field is essentially uniform across her head as her head is relatively very small as compared the distance from the black hole. The minute difference due to slight non-uniformity in the gravitational pull is not enough to displace the center of gravity far away from the center of mass.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravitational Force
Gravitational force is a fundamental force that attracts any two objects with mass. The strength of this force depends on the masses involved and the distance separating them. The formula to calculate gravitational force is given by \( F = \frac{G M m}{r^2} \), where:
  • \( G \) is the gravitational constant \(~6.674 脳 10^{-11} \text{Nm}^2/\text{kg}^2\).
  • \( M \) and \( m \) represent the masses of the two objects.
  • \( r \) is the distance between the centers of the two masses.
In the context of the exercise, we calculate the gravitational force acting on the astronaut and her ears, which allows us to assess the variations in gravitational pull, known as tidal forces. These variations occur due to the black hole's immense gravity, creating significant differences in force along the astronaut's body.
Schwarzschild Radius
The Schwarzschild radius represents a key concept in understanding black holes. It defines the boundary beyond which nothing, not even light, can escape the gravitational pull of a black hole. The radius depends solely on the mass of the black hole, calculated by the formula: \[ r_s = \frac{2GM}{c^2} \]where:
  • \( G \) is the gravitational constant.
  • \( M \) is the mass of the black hole.
  • \( c \) is the speed of light in a vacuum \(~3.00 脳 10^8 \text{m/s}\).
For the black hole in the exercise, being 5 times the mass of the Sun, the Schwarzschild radius is given as 15 km. This distance is crucial for determining the boundary of the black hole's "event horizon," beyond which objects cannot escape its gravitational grip.
Center of Mass
The center of mass of an object is essentially the point where its mass is evenly distributed and balanced. It's a pivotal concept in physics because it simplifies the analysis of motion, as you can model the movement of the object as if all of its mass were concentrated at this point.
For the astronaut in the exercise, the center of mass is important because it helps in understanding how her body would orbit around the black hole. Her body, ears included, rotates around this center of mass as she orbits the black hole, despite the tidal forces trying to stretch her.
Center of Gravity
The center of gravity is often considered equivalent to the center of mass. However, it relates specifically to how gravitational forces act on an object. In a uniform gravitational field, both the center of mass and the center of gravity coincide.
In the exercise, even though the gravitational field is stronger at the point closer to the black hole, the small size of the astronaut's head compared to the distance from the black hole means that the difference in gravitational pull is negligible. Thus, her center of gravity remains almost the same as her center of mass.
Orbital Dynamics
Orbital dynamics studies how objects move in their orbits, influenced by gravitational forces. Key factors include velocity, distance from the central body, and the resulting gravitational forces.
For the astronaut orbiting the black hole, her entire body moves with the same angular velocity. This means her body parts should technically stay in orbit unless acted upon by additional forces, like the tidal forces in the exercise.
  • Her ears experience different gravitational pull due to their unique positions, leading to varying velocities along the orbit.
  • This creates internal forces exerted by her head to keep her ears in the correct orbit.
Understanding these dynamics helps assess whether she would be torn apart due to the differing forces acting upon her in such a dramatic gravitational environment.

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Most popular questions from this chapter

A thin spherical shell has radius \(r_{A}=4.00 \mathrm{~m}\) and mass \(m_{A}=20.0 \mathrm{~kg} .\) It is concentric with a second thin spherical shell that has radius \(r_{B}=6.00 \mathrm{~m}\) and mass \(m_{B}=40.0 \mathrm{~kg} .\) What is the net gravitational force that the two shells exert on a point mass of \(0.0200 \mathrm{~kg}\) that is a distance \(r\) from the common center of the two shells, for (a) \(r=2.00 \mathrm{~m}\) (inside both shells), (b) \(r=5.00 \mathrm{~m}\) (in the space between the two shells), and (c) \(r=8.00 \mathrm{~m}\) (outside both shells)?

Rhea, one of Saturn's moons, has a radius of \(764 \mathrm{~km}\) and an acceleration due to gravity of \(0.265 \mathrm{~m} / \mathrm{s}^{2}\) at its surface. Calculate its mass and average density.

DATA For a spherical planet with mass \(M,\) volume \(V,\) and radius \(R,\) derive an expression for the acceleration due to gravity at the planet's surface, \(g\), in terms of the average density of the planet, \(\rho=M / V,\) and the planet's diameter, \(D=2 R .\) The table gives the values of \(D\) and \(g\) for the eight major planets: $$ \begin{array}{lrc} \text { Planet } & D(\mathrm{~km}) & g\left(\mathrm{~m} / \mathrm{s}^{2}\right) \\ \hline \text { Mercury } & 4879 & 3.7 \\ \text { Venus } & 12,104 & 8.9 \\ \text { Earth } & 12,756 & 9.8 \\ \text { Mars } & 6792 & 3.7 \\ \text { Jupiter } & 142,984 & 23.1 \\ \text { Saturn } & 120,536 & 9.0 \\ \text { Uranus } & 51,118 & 8.7 \\ \text { Neptune } & 49.528 & 11.0 \end{array} $$ (a) Treat the planets as spheres. Your equation for \(g\) as a function of \(\rho\) and \(D\) shows that if the average density of the planets is constant, a graph of \(g\) versus \(D\) will be well represented by a straight line. Graph 8 as a function of \(D\) for the eight major planets. What does the graph tell you about the variation in average density? (b) Calculate the average density for each major planet. List the planets in order of decreasing density, and give the calculated average density of each. (c) The earth is not a uniform sphere and has greater density near its center. It is reasonable to assume this might be true for the other planets. Discuss the effect this has on your analysis. (d) If Saturn had the same average density as the earth, what would be the value of \(g\) at Saturn's surface?

Two spherically symmetric planets with no atmosphere have the same average density, but planet \(B\) has twice the radius of planet \(A\). A small satellite of mass \(m_{A}\) has period \(T_{A}\) when it orbits planet \(A\) in a circular orbit that is just above the surface of the planet. A small satellite of mass \(m_{B}\) has period \(T_{B}\) when it orbits planet \(B\) in a circular orbit that is just above the surface of the planet. How does \(T_{B}\) compare to \(T_{A} ?\) with a density of \(2500 \mathrm{~kg} / \mathrm{m}^{3} ?\)

A narrow uniform rod has length \(2 a\). The linear mass density of the rod is \(\rho,\) so the mass \(m\) of a length \(l\) of the rod is \(\rho l\). (a) A point mass is located a perpendicular distance \(r\) from the center of the rod. Calculate the magnitude and direction of the force that the rod exerts on the point mass. (Hint: Let the rod be along the \(y\) -axis with the center of the rod at the origin, and divide the rod into infinitesimal segments that have length \(d y\) and that are located at coordinate \(y\). The mass of the segment is \(d m=\rho d y\). Write expressions for the \(x\) - and \(y\) -components of the force on the point mass, and integrate from \(-a\) to \(+a\) to find the components of the total force. Use the integrals in Appendix B.) (b) What does your result become for \(a \gg r ?\) (Hint: Use the power series for \((1+x)^{n}\) given in Appendix B.) (c) For \(a \gg r,\) what is the gravitational field \(g=\boldsymbol{F}_{g} / m\) at a distance \(r\) from the rod? (d) Consider a cylinder of radius \(r\) and length \(L\) whose axis is along the rod. As in part (c), let the length of the rod be much greater than both the radius and length of the cylinder. Then the gravitational ficld is constant on the curved side of the cylinder and perpendicular to it, so the gravitational flux \(\Phi_{g}\) through this surface is cqual to \(g A\), where \(A=2 \pi r L\) is the area of the curved side of the cylinder (see Problem 13.59 ). Calculate this flux. Write your result in terms of the mass \(M\) of the portion of the rod that is inside the cylindrical surface. How does your result depend on the radius of the cylindrical surface?
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