/*! 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 39 Suppose that on earth you can ju... [FREE SOLUTION] | 91Ó°ÊÓ

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Suppose that on earth you can jump straight up a distance of \(50 \mathrm{cm} .\) Can you escape from a \(4.0-\mathrm{km}\) -diameter asteroid with a mass of \(1.0 \times 10^{14} \mathrm{kg} ?\)

Short Answer

Expert verified
Whether one can escape from an asteroid of described characteristics based on their jumping ability on Earth can be determined by comparing the gravitational force on the asteroid and the force needed to jump to a given height on Earth.

Step by step solution

01

Calculation of Gravitational Force

Firstly, use Newton's Law of Universal Gravitation to calculate the gravitational force (\(F_{g}\)) on the asteroid. The formula is \(F_{g} = \frac{G m_{1}m_{2}}{r^{2}}\) where \(G = 6.674 \times 10^{-11} \, m^{3}kg^{-1}s^{-2} \) is the universal gravitational constant, \(m_{1}\) and \(m_{2}\) are the masses of two bodies (the mass of the person on earth is approximately 70 kg), and \(r\) is the distance between the object and the center of the asteroid (half its diameter).
02

Calculation of Force Needed to Jump

Secondly, we calculate the force (\(F_{jump}\)) needed to jump 50cm (or 0.5m) on earth. This requires Newton's Second Law, \( F = ma\), where \(m\) is the mass of the person and \(a\) is acceleration due to gravity on earth (9.8 m/s^2). Thus, \(F_{jump} = m \cdot a\). This is the force applied to jump on earth.
03

Comparison of Forces

Finally, the two forces are compared. If the force needed to jump on the asteroid is less than on earth (\(F_{jump} > F_{g}\)), then the person can escape from the asteroid, as they have enough force to overcome the asteroid's gravitational pull.

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Key Concepts

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

Newton's Law of Universal Gravitation
Newton's Law of Universal Gravitation is the cornerstone in understanding how objects attract each other due to their masses. This universal force governs everything from the apples falling from trees to the orbits of moons around planets. The formula is expressed as \[ F_{g} = \frac{G m_{1}m_{2}}{r^{2}} \]In this equation:
  • \( F_{g} \) represents the gravitational force.
  • \( G \) is the gravitational constant.
  • \( m_{1} \) and \( m_{2} \) are the masses of the two attracting bodies.
  • \( r \) is the distance between the centers of the two masses.
This equation shows that the force of gravity decreases with the square of the distance, meaning objects further apart exert less pull on each other. This principle helps us predict how strong gravitational interactions are depending on distance and mass, whether it's jumping off an asteroid or how planets move in space.
Understanding this law gives us insight into the balance of forces that keep our universe in motion.
Gravitational constant (G)
The gravitational constant, denoted as \( G \), is a critical component in calculating gravitational forces. It is a universal constant, meaning it has the same value all over the universe. In the formula \( F_{g} = \frac{G m_{1}m_{2}}{r^{2}} \), \( G \) provides the necessary proportionality to the gravitational equation, allowing us to quantify the force between two masses.The value of \( G \) is approximately \( 6.674 \times 10^{-11} \, m^{3}kg^{-1}s^{-2} \). It is quite small, which explains why we don't feel the gravitational forces between small everyday objects. This small number indicates that only very massive objects, like planets and stars, produce significant gravitational forces.
If you calculate the gravitational force without \( G \), the results would not make physical sense, as the units and magnitudes would be incorrect. \( G \) is essential for making theoretical calculations match our observations of real-world phenomena, allowing us to understand how massive objects influence each other.
Acceleration due to gravity
Acceleration due to gravity is an important concept that describes how quickly an object will accelerate towards the center of a mass, like a planet or asteroid. On Earth, this acceleration is approximately \( 9.8 \, m/s^2 \). This means that for every second an object is in freefall, its downward velocity increases by \( 9.8 \, m/s \).While \( 9.8 \, m/s^2 \) is typical for Earth, other celestial bodies like the moon or asteroids will have different values of gravitational acceleration. The acceleration depends on the mass of the body and the radius (or distance from the center to the surface).When considering whether you can escape an asteroid's pull, this acceleration affects how much force you need to overcome that pull. If the acceleration is low, like in the case of a small asteroid, the force required to jump off may be feasible with a normal human jump. This is in contrast to Earth, where the higher acceleration means more force is necessary to leap upwards. Understanding gravity’s acceleration helps explain everything from how space missions navigate to why things fall at certain speeds.

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Most popular questions from this chapter

In Problems 65 through 68 you are given the equation(s) used to solve a problem. For each of these, you are to a. Write a realistic problem for which this is the correct equation(s). b. Draw a pictorial representation. c. Finish the solution of the problem. $$\begin{array}{l} \left(2.0 \times 10^{30} \mathrm{kg}\right) \mathrm{v}_{\mathrm{n}}+\left(4.0 \times 10^{30} \mathrm{kg}\right) v_{0}=0 \\ \frac{1}{2}\left(2.0 \times 10^{30} \mathrm{kg}\right) \mathrm{v}_{\mathrm{n}}^{2}+\frac{1}{2}\left(4.0 \times 10^{30} \mathrm{kg}\right) \mathrm{v}_{\mathrm{R}}^{2} \\ -\frac{\left(6.67 \times 10^{-11} \mathrm{Nm}^{2} / \mathrm{kg}^{2}\right)\left(2.0 \times 10^{30} \mathrm{kg}\right)\left(4.0 \times 10^{30} \mathrm{kg}\right)}{1.0 \times 10^{9} \mathrm{m}} \\ =0+0 \\ -\frac{\left(6.67 \times 10^{-11} \mathrm{Nm}^{2} / \mathrm{kg}^{2}\right)\left(2.0 \times 10^{10} \mathrm{kg}\right)\left(4.0 \times 10^{20} \mathrm{kg}\right)}{1.0 \times 10^{12} \mathrm{m}} \end{array}$$

A huge cannon is assembled on an airless planet. The planet has a radius of \(5.0 \times 10^{4} \mathrm{m}\) and a mass of \(2.6 \times 10^{24} \mathrm{kg}\). The cannon fires a projectile straight up at \(5000 \mathrm{m} / \mathrm{s}\) a. What height does the projectile reach above the surface? b. An observation satellite orbits the planet at a height of \(1000 \mathrm{km} .\) What is the projectile's speed as it passes the satellite?

The space shuttle is in a \(250-\mathrm{km}\) -high circular orbit. It needs to reach a \(610-\mathrm{km}\) -high circular orbit to catch the Hubble Space Telescope for repairs. The shuttle's mass is 75,000 kg. How much energy is required to boost it to the new orbit?

In Problems 65 through 68 you are given the equation(s) used to solve a problem. For each of these, you are to a. Write a realistic problem for which this is the correct equation(s). b. Draw a pictorial representation. c. Finish the solution of the problem. $$\begin{array}{l} \frac{\left(6.67 \times 10^{-11} \mathrm{Nm}^{2} / \mathrm{kg}^{2}\right)\left(5.68 \times 10^{26} \mathrm{kg}\right)}{r^{2}} \\ =\frac{\left(6.67 \times 10^{-11} \mathrm{Nm}^{2} / \mathrm{kg}^{2}\right)\left(5.98 \times 10^{24} \mathrm{kg}\right)}{\left(6.37 \times 10^{6} \mathrm{m}\right)^{2}} \end{array}$$

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