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An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030-kg ears is 6.0 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 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 ears 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
(a) The tension is small; she won't be torn apart. (b) Center of gravity and center of mass coincide.

Step by step solution

01

Understand the Problem

We need to evaluate the gravitational effect on the astronaut's ears, resulting in a gravitational force, which creates a tension between the two positions due to their differing distances from the black hole. We also need to determine whether the center of gravity and center of mass align in her head.
02

Calculating Gravitational Force Difference

Using the formula for gravitational force: \[ F = \frac{G M m}{r^2} \]where \(G\) is the gravitational constant, \(M\) is the mass of the black hole, \(m\) is the mass of one ear, and \(r\) is the radius of orbit.First, calculate the force on the ear closer to the black hole \( F_1 \) at distance \( 120\, \text{km} - 6\, \text{cm} \), converting cm to km.Next, calculate the force on the ear farther from the black hole \( F_2 \) at distance \( 120\, \text{km} + 6\, \text{cm} \).The tension \( T \) is \( T = F_2 - F_1 \).
03

Mass of the Black Hole

The mass of the black hole is 5.00 times the mass of the sun. Using the sun's mass, \( M_{\odot} = 1.989 \times 10^{30}\, \text{kg} \), the mass of the black hole \( M = 5.00 \times M_{\odot} = 9.945 \times 10^{30}\, \text{kg} \).
04

Calculate Forces on Ears

For the closer ear, the radius is 119.994 km; for the further ear, it is 120.006 km. Using \( G = 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \), calculate \( F_1 \) and \( F_2 \) using the formula \[ F = \frac{(6.674 \times 10^{-11}) \times (9.945 \times 10^{30}) \times 0.030}{r^2} \] Compute \( F_1 \) and \( F_2 \) for their respective radii.
05

Calculate Tension Between Ears

Now, calculate the difference between the forces: \[ T = F_2 - F_1 \] This gives the tension between the ears due to differing gravitational forces.
06

Determine Difficulty in Maintaining Integrity

Based on the tension calculated, consider whether this tension is significant enough to potentially tear apart the astronaut, considering typical material strength and biological limits.
07

Evaluate Center of Gravity vs Center of Mass

Center of gravity and center of mass are effectively at the same point if the gravitational field is uniform. Given the small distance compared to the black hole's scale, the gravitational field over her head can be considered nearly uniform. Thus, there's no significant difference between them in this context.

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Key Concepts

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

Tidal Forces
Tidal forces are fascinating phenomena that occur due to the variation in gravitational strength over an object. They are particularly important in our original exercise, where the astronaut feels different gravitational pulls on her ears.
  • The closer ear experiences a stronger force, while the farther ear feels a weaker one.
  • This difference creates a tension known as tidal force, which in extreme cases, can stretch or even tear apart objects (a process humorously termed "spaghettification").
In general, tidal forces arise when an object is large enough that different parts of it experience varying gravitational fields. The larger the distance between two points on the object, the more pronounced the tidal force becomes. In our case, with the astronaut only 6 cm apart on either side of her head, the forces light up subtle but important aspects of gravitational physics related to black holes.
Schwarzschild Radius
The Schwarzschild radius defines the boundary of a black hole, where its gravitational pull becomes so strong that not even light can escape.
  • This concept helps us understand the true nature and boundary of black holes.
  • It's calculated using the formula: \[ R_s = \frac{2GM}{c^2} \] where \( G \) is the gravitational constant, \( M \) is the mass, and \( c \) is the speed of light.
For example, in our exercise, the Schwarzschild radius of the black hole is 15 km, making it the point beyond which nothing can return. The astronaut orbits outside this radius, ensuring she remains in the safe zone, at least for the time being. Understanding this concept is critical when dealing with black holes, as it sets the stage for discussing event horizons and the singularity at a black hole's core.
Center of Gravity vs Center of Mass
The center of gravity and center of mass are often considered synonyms, but they have subtle differences. Both are crucial in physics to understand the behavior of objects in gravitational fields. In straightforward cases like our exercise:
  • The center of mass is a point representing the average position of all the mass in an object.
  • The center of gravity is the point where gravitational forces act as if they were concentrated.
In a uniform gravitational field, these points coincide. Since the gravitational field change over the astronaut’s small head is negligible, both centers align in her case. Understanding this equivalence helps clarify calculations in physics, ensuring that our estimates of gravitational impacts are both precise and consistent. This concept makes analyzing more complex scenarios involving gravity easier by simplifying the geometry involved.

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Most popular questions from this chapter

Planet Vulcan. Suppose that a planet were discovered between the sun and Mercury, with a circular orbit of radius equal to \(\frac{2}{3}\) of the average orbit radius of Mercury. What would be the orbital period of such a planet? (Such a planet was once postulated, in part to explain the precession of Mercury's orbit. It was even given the name Vulcan, although we now have no evidence that it actually exists. Mercury's precession has been explained by general relativity.)

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