/*! 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 43 What is the free-fall accelerati... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

What is the free-fall acceleration at the surface of (a) Mars and (b) Jupiter?

Short Answer

Expert verified
The free-fall acceleration at the surface of (a) Mars is approximately \( 3.71 m/s^2 \) and (b) Jupiter is approximately \( 24.79 m/s^2 \).

Step by step solution

01

Gather the needed values

Before any calculations can take place, the needed values must be gathered. The gravitational constant \( G \) stays the same, being \( 6.674 \times 10^{-11} m^3 kg^{-1} s^{-2} \). For the mass \( M \) and radius \( r \) of Mars, these values are respectively \( 6.39 \times 10^{23} kg \) and \( 3.39 \times 10^{6} m \). For the mass and radius of Jupiter, these values are respectively \( 1.898 \times 10^{27} kg \) and \( 6.9911 \times 10^{7} m \).
02

Calculate the free-fall acceleration of Mars

Using the gathered values, the formula \( g = \frac{GM}{r^2} \) can be solved. Substituting Mars' values into the formula, the calculation of \( g \) for Mars becomes \( g = \frac{(6.674 \times 10^{-11} m^3 kg^{-1} s^{-2})(6.39 \times 10^{23} kg)}{(3.39 \times 10^{6} m)^2} \). Evaluating this gives a value of approximately \( 3.71 m/s^2 \) for the free-fall acceleration on Mars' surface.
03

Calculate the free-fall acceleration of Jupiter

In the same manner as to calculate Mars' \( g \), substitute in Jupiter's values into the formula \( g = \frac{GM}{r^2} \). This yields \( g = \frac{(6.674 \times 10^{-11} m^3 kg^{-1} s^{-2})(1.898 \times 10^{27} kg)}{(6.9911 \times 10^{7} m)^2} \). Evaluating this provides the result of approximately \( 24.79 m/s^2 \) for the free fall acceleration at Jupiter's surface.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Gravitational Constant
Understanding the gravitational constant (\(G\)) is fundamental to grasping the concept of free-fall acceleration on celestial bodies. It's a universal value that determines the force of gravity's pull between two masses. Specifically, the gravitational constant is a fixed number, approximately equal to \(6.674 \times 10^{-11} m^3 kg^{-1} s^{-2}\), which enters into the calculation of gravitational force. It partakes in determining the gravitational attraction regardless of the masses or the distance involved.

When calculating free-fall acceleration on a planet, \(G\) is utilized as part of Newton's law of universal gravitation, which predicts how objects with mass attract one another. It's this constancy that enables us to compare gravitational forces on different planets as long as we have accurate measurements of their masses and radii.
Planetary Mass
The mass of a planet, often denoted by the symbol \(M\), is a measure of the amount of matter it contains. This value is extremely important when calculating gravitational forces. Since gravity is directly dependent on mass, heavier planets exert a greater gravitational pull than lighter ones.

This explains why, for instance, the free-fall acceleration on Jupiter is considerably higher than on Mars. Jupiter's mass is roughly 317 times that of Earth, while Mars is only about one-tenth of Earth's mass. The enormous mass of Jupiter, at \(1.898 \times 10^{27} kg\), results in a stronger gravitational force exerted on objects at the surface compared to Mars, which has a mass of \(6.39 \times 10^{23} kg\). Consequently, understanding a planet's mass is crucial in predicting the force of gravity an object will experience on that planet.
Planetary Radius
The radius of a planet (\(r\)) plays a crucial role in determining the intensity of its gravitational field at the surface. The radius is the distance from the planet's center to its surface. An increase in the planetary radius, while mass remains constant, would result in a weaker gravitational pull at its surface.

Since the formula for calculating free-fall acceleration includes the radius in the denominator, squared (\(r^2\)), a larger radius significantly diminishes the acceleration due to gravity. This relationship is vivid when comparing planets like Mars and Jupiter; despite the latter’s greater mass, its larger radius of \(6.9911 \times 10^{7} m\) compared to Mars' \(3.39 \times 10^{6} m\) influences the calculation. These measurements are essential to accurately determine the gravitational force experienced on the surface of assorted celestial bodies.
Newton’s Law of Universal Gravitation
Newton's law of universal gravitation is a cornerstone in physics, providing an equation that relates the gravitational attraction between two bodies to their masses and the distance between them. It is mathematically expressed as \(F = G\frac{Mm}{r^2}\), where \(F\) is the gravitational force, \(G\) is the gravitational constant, \(M\) and \(m\) are the masses of two objects, and \(r\) is the distance between the centers of the two masses.

This law is imperative when determining the free-fall acceleration of objects on planets. To find the free-fall acceleration 'g' at a planet's surface, the formula simplifies to \(g = \frac{GM}{r^2}\), as the object's mass (m) cancels out. The direct application of this principle allows us to compute the free-fall accelerations on Mars and Jupiter as done in the textbook exercise. Newton's law illustrates why heavier planets (higher \(M\)) or smaller radii (lower \(r\)) would lead to greater free-fall accelerations, while the converse would yield weaker gravitational pull.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A fast pitch softball player does a "windmill" pitch, illustrated in Figure \(\mathrm{P} 6.18\), moving her hand through a circular arc to pitch a ball at 70 mph. The 0.19 kg ball is 50 cm from the pivot point at her shoulder. At the lowest point of the circle, the ball has reached its maximum speed. a. At the bottom of the circle, just before the ball leaves her hand, what is its centripetal acceleration? b. What are the magnitude and direction of the force her hand exerts on the ball at this point?

In recent years, scientists have discovered hundreds of planets orbiting other stars. Some of these planets are in orbits that are similar to that of earth, which orbits the sun \(\left(M_{\text {sun }}=1.99 \times 10^{30} \mathrm{kg}\right)\) at a distance of \(1.50 \times 10^{11} \mathrm{m},\) called 1 astronomical unit \((1\) au \() .\) Others have extreme orbits that are much different from anything in our solar system. Problems \(51-53\) relate to some of these planets that follow circular orbits around other stars. HD \(10180 \mathrm{g}\) orbits with a period of 600 days at a distance of 1.4 au from its star. What is the ratio of the star's mass to our sun's mass?

The horse on a carousel is \(4.0 \mathrm{m}\) from the central axis. a. If the carousel rotates at 0.10 rev/s, how long does it take the horse to go around twice? b. How fast is a child on the horse going (in \(\mathrm{m} / \mathrm{s}\) )?

In a recent study of how mice negotiate turns, the mice ran around a circular \(90^{\circ}\) turn on a track with a radius of \(0.15 \mathrm{m} .\) The maximum speed measured for a mouse (mass \(=18.5\) g \()\) running around this turn was \(1.29 \mathrm{m} / \mathrm{s}\). What is the minimum coefficient of friction between the track and the mouse's feet that would allow a turn at this speed?

\(\mathrm{A} 1500 \mathrm{kg}\) car drives around a flat \(200-\mathrm{m}\) -diameter circular track at \(25 \mathrm{m} / \mathrm{s}\). What are the magnitude and direction of the net force on the car? What causes this force?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.