/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 88 Tidal Forces near a Black Hole. ... [FREE SOLUTION] | 91Ó°ÊÓ

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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 -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 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
The tension is negligible; the astronaut won't be torn apart. Center of gravity aligns with center of mass.

Step by step solution

01

Determine the Gravitational Force Formula

The gravitational force exerted on an object by a massive body is given by the formula:\[ F = \frac{G \, M \, m}{r^2} \]where \( F \) is the force, \( G \) is the universal gravitational constant \( 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \), \( M \) is the mass of the black hole, \( m \) is the mass of the object, and \( r \) is the distance from the center of the massive body to the object.
02

Calculate Mass of the Black Hole

The mass of the sun \( M_\odot \) is approximately \( 1.989 \times 10^{30} \, \text{kg} \). The black hole has a mass 5.00 times that of the sun, so:\[ M = 5.00 \times 1.989 \times 10^{30} \, \text{kg} = 9.945 \times 10^{30} \, \text{kg} \]
03

Calculate the Gravitational Force on Each Ear

The distance from the black hole to each ear differs by \( 6.0 \, \text{cm} \). For the ear closer to the black hole, the distance is \( 120 \, \text{km} - 6.0 \, \text{cm} \) and for the farther ear, it is \( 120 \, \text{km} + 6.0 \, \text{cm} \). Convert these to meters:- Closer ear: \( r_1 = 120,000 \, \text{m} - 0.06 \, \text{m} = 119,999.94 \, \text{m} \)- Farther ear: \( r_2 = 120,000 \, \text{m} + 0.06 \, \text{m} = 120,000.06 \, \text{m} \).Calculate the forces:\[ F_1 = \frac{G \times 9.945 \times 10^{30} \, \text{kg} \times 0.030 \, \text{kg}}{(119,999.94 \, \text{m})^2} \]\[ F_2 = \frac{G \times 9.945 \times 10^{30} \, \text{kg} \times 0.030 \, \text{kg}}{(120,000.06 \, \text{m})^2} \]
04

Calculate the Difference in Force (Tension)

The tension between the ears (which is the difference in gravitational force) is:\[ \Delta F = F_2 - F_1 \]Calculating this using the previously outlined values will give the tension the astronaut feels.
05

Addressing the Potential for Harm

Given the small mass of 0.030 kg (30 grams) and the tiny difference in forces (due to the small difference in radius), the resulting tension is extremely small. Thus, the astronaut would not be torn apart easily by these tidal forces near the black hole at the given distance.
06

Discussing Center of Gravity vs. Center of Mass

The center of gravity is dependent on the gravitational field and the distribution of mass. However, in a uniform gravitational field like on Earth or in orbit, the center of gravity coincides with the center of mass. Hence, they are effectively the same for practical purposes in this scenario.

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Key Concepts

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

Black Holes
Black holes are fascinating cosmic objects with such strong gravitational pulls that nothing, not even light, can escape from them.
They are formed when a massive star collapses under its own gravity, transforming into an extremely dense point known as a singularity.
  • Event Horizon: This is the point of no return. If something crosses this boundary, it cannot escape the black hole's gravity.

  • Singularity: The center of a black hole where density becomes infinite and gravitational forces are extremely intense.

  • Strong Gravitational Pull: Black holes can have masses millions of times that of our sun, exerting immense gravitational forces.
When dealing with black holes, tidal forces come into play. These forces arise because gravitational pull varies significantly from one part of an object, like a spaceship, to another when they are near a massive body like a black hole.
This exercise explores these forces and how they affect objects in close proximity to a black hole.
Gravitational Force
Gravitational force is a natural phenomenon by which objects with mass attract one another.
It governs the motion of celestial bodies and keeps our feet planted firmly on the ground. This force is described mathematically by Newton's Law of Universal Gravitation.
  • Formula: \[ F = \frac{G \times M \times m}{r^2} \] where:
    • \( F \) is the gravitational force,

    • \( G \) is the universal gravitational constant,

    • \( M \) and \( m \) are the masses of two bodies, and

    • \( r \) is the distance between their centers.
  • Universal Gravitational Constant (\( G \)): This constant helps calculate gravitational forces between objects. It's value is \( 6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \).
The gravitational force is influenced by the mass of both interacting objects and the square of the distance between them.
As the distance increases, the gravitational force decreases rapidly. In our scenario, this explains why there is a small difference in force between the astronaut's ears, even near a massive black hole.
Schwarzschild Radius
The Schwarzschild radius is a measure that defines the size of the event horizon of a black hole.
It represents how compact the mass of an object must be for it to become a black hole.
  • Calculation: Determined using the formula: \[ r_s = \frac{2G M}{c^2} \] where:
    • \( r_s \) is the Schwarzschild radius,

    • \( G \) is the gravitational constant,

    • \( M \) is the mass of the object, and

    • \( c \) is the speed of light.
  • Relation to Mass: The greater the mass of the object, the larger its Schwarzschild radius.
For massive objects like black holes, the Schwarzschild radius is quite significant as it helps define the boundary beyond which the object's gravitational pull becomes strong enough to prevent anything from escaping.
In this exercise, the black hole has a Schwarzschild radius of 15 km, indicating the size of its event horizon.

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Most popular questions from this chapter

Planet \(\mathrm{X}\) rotates in the same manner as the earth, around an axis through its north and south poles, and is perfectly spherical. An astronaut who weighs 943.0 \(\mathrm{N}\) on the earth weighs 915.0 \(\mathrm{N}\) at the north pole of Planet \(\mathrm{X}\) and only 850.0 \(\mathrm{N}\) at its equator. The distance from the north pole to the equator is \(18,850 \mathrm{km},\) measured along the surface of Planet \(\mathrm{X}\) . (a) How long is the day on Planet \(\mathrm{X}\) ? (b) If a 45,000-kg satellite is placed in a circular orbit 2000 \(\mathrm{km}\) above the surface of Planet \(\mathrm{X},\) what will be its orbital period?

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

Cavendish Experiment. In the Cavendish balance apparatus shown in Fig. \(13.4,\) suppose that \(m_{1}=1.10 \mathrm{kg}, m_{2}=\) \(25.0 \mathrm{kg},\) and the rod connecting the \(m_{1}\) pairs is 30.0 \(\mathrm{cm}\) long. If, in each pair, \(m_{1}\) and \(m_{2}\) are 12.0 \(\mathrm{cm}\) apart center to center, find (a) the net force and (b) the net torque (about the rotation axis) on the rotating part of the apparatus. (c) Does it seem that the torque in part (b) would be enough to easily rotate the rod? Suggest some ways to improve the sensitivity of this experiment.

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