/*! 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 16 The mass of Venus is 81.5\(\%\) ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The mass of Venus is 81.5\(\%\) that of the earth, and its radius is 94.9\(\%\) that of the earth. (a) Compute the acceleration due to gravity on the surface of Venus from these data. (b) If a rock weighs 75.0 \(\mathrm{N}\) on earth, what would it weigh at the surface of Venus?

Short Answer

Expert verified
(a) The acceleration due to gravity on Venus is approximately 8.87 m/s². (b) The rock would weigh about 67.8 N on Venus.

Step by step solution

01

Identify the Earth's Gravitational Formula

The acceleration due to gravity on the surface of a planet can be calculated using the formula: \( g = \frac{G \cdot M}{R^2} \), where \( g \) is the acceleration due to gravity, \( G \) is the universal gravitational constant \( (6.674 \times 10^{-11} \mathrm{Nm}^2/\mathrm{kg}^2) \), \( M \) is the mass of the planet, and \( R \) is the radius of the planet. For Earth, \( g = 9.81\, \mathrm{m/s}^2 \).
02

Determine Venus' Mass and Radius

Given that the mass of Venus is 81.5\( \% \) of Earth's mass and the radius is 94.9\( \% \) of Earth's radius, express these as fractions of Earth's values: \( M_{\text{Venus}} = 0.815 M_{\text{Earth}} \) and \( R_{\text{Venus}} = 0.949 R_{\text{Earth}} \).
03

Express Venus' Gravitational Acceleration

Using the derived expressions, substitute into the gravity formula: \( g_{\text{Venus}} = \frac{G \cdot 0.815 M_{\text{Earth}}}{(0.949 R_{\text{Earth}})^2} \). This simplifies to \( g_{\text{Venus}} = \frac{0.815}{0.949^2} g_{\text{Earth}} \).
04

Compute the Gravitational Acceleration on Venus

Now, calculate \( g_{\text{Venus}} \) using Earth's \( g \): \( g_{\text{Venus}} = \frac{0.815}{0.949^2} \times 9.81 \approx 8.87 \, \mathrm{m/s}^2 \).
05

Determine Weight on Venus Surface

Weight is the force due to gravity, calculated by \( F = m \cdot g \). If a rock weighs 75.0 \( \mathrm{N} \) on Earth, its mass \( m \) is \( \frac{75}{9.81} \approx 7.645 \mathrm{kg} \).
06

Calculate Rock's Weight on Venus

Using Venus' gravitational acceleration, the rock's weight on Venus is: \( F_{\text{Venus}} = 7.645 \times 8.87 \approx 67.8 \mathrm{N} \).

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.

Mass of Venus
The mass of Venus is a crucial factor in determining the gravitational pull that the planet exerts. Venus, sometimes called Earth’s "sister planet" due to its similar size and composition, has a mass that is approximately 81.5% of Earth's mass. The mass of a planet significantly impacts how much force it can exert on objects at its surface.

In this exercise, we translate the percentage of Earth's mass that Venus has into an algebraic expression. If Earth’s mass is denoted by \( M_{\text{Earth}} \), then the mass of Venus is expressed as \( M_{\text{Venus}} = 0.815 \times M_{\text{Earth}} \).

This value is essential because the gravitational force depends directly on the planetary mass. More mass means more gravitational pull.
Radius of Venus
Venus doesn't just have less mass than Earth, but it also has a slightly smaller radius. The radius of a planet is the distance from its center to its surface, and for Venus, this is about 94.9% that of Earth's radius. This percentage allows us to calculate Venus' radius as \( R_{\text{Venus}} = 0.949 \times R_{\text{Earth}} \).

When using this factor in gravitational calculations, it is important to remember that the radius appears squared in the formula for gravitational acceleration. This means even small differences in radius can significantly affect gravitational pull.
  • The radius affects the gravitational pull inversely - as the radius increases, gravitational pull decreases when mass stays constant.
  • In the formula \( g = \frac{G \cdot M}{R^2} \), the radius is squared, so its impact is quite profound.
Weight on Different Planets
Weight varies depending on the planet due to differences in gravitational pull. Essentially, weight is the force of gravity acting on an object's mass. On Earth, a rock weighing 75 Newtons is an example often used to illustrate this concept. But how does this change on Venus?

The weight of the rock changes because the gravity on Venus is different. Assuming the rock's mass remains constant, its weight on Venus is calculated using Venus' gravitational acceleration, which is roughly 8.87 m/s².
  • We first find the rock's mass by dividing its weight on Earth by Earth's gravity: \( m = \frac{75}{9.81} \text{ kg} \).
  • We can then find its weight on Venus by multiplying its mass by Venus' gravitational acceleration, resulting in approximately 67.8 Newtons.
Notice that the weight is less on Venus, illustrating how gravitational force impacts weight.
Universal Gravitational Constant
The Universal Gravitational Constant, denoted as \( G \), is a fundamental component in the law of universal gravitation. It is a constant value that describes the strength of gravity in Newton's Law of Universal Gravitation. The value of \( G \) is approximately \( 6.674 \times 10^{-11} \text{ Nm}^2/\text{kg}^2 \).

This constant allows us to calculate the gravitational force between two masses. Crucially, it stays the same irrespective of the specific masses or the distance between them. This concept helps to simplify calculations while ensuring accuracy across varying conditions in space.
  • \( G \) is pivotal in the gravity formula: \( g = \frac{G \cdot M}{R^2} \).
  • It does not change over different planets; instead, the mass \( M \) and radius \( R \) of the planet do, altering \( g \), the acceleration due to gravity.
Understanding \( G \) is essential for computations involving any celestial body, not just Venus.

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

In 2005 astronomers announced the discovery of a large black hole in the galaxy Markarian 766 having clumps of matter orbiting around once every 27 hours and moving at \(30,000 \mathrm{km} / \mathrm{s}\) . (a) How far are these clumps from the center of the black hole? (b) What is the mass of this black hole, assuming circular orbits? Express your answer in kilograms and as a multiple of our sun's mass. (c) What is the radius of its event horizon?

(a) Asteroids have average densities of about 2500 \(\mathrm{kg} / \mathrm{m}^{3}\) and radii from 470 \(\mathrm{km}\) down to less than a kilometer. Assuming that the asteroid has a spherically symmetric mass distribution, estimate the radius of the largest asteroid from which you could escape simply by jumping off. (Hint: You can estimate your jump speed by relating it to the maximum height that you can jump on earth. (b) Europa, one of Jupiter's four large moons, has a radius of 1570 \(\mathrm{km}\) . The acceleration due to gravity at its surface is 1.33 \(\mathrm{m} / \mathrm{s}^{2}\) . Calculate its average density.

An \(8.00-\mathrm{kg}\) point mass and a \(15.0-\mathrm{kg}\) 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.

A uniform sphere with mass 60.0 \(\mathrm{kg}\) is held with its center at the origin, and a second uniform sphere with mass 80.0 \(\mathrm{kg}\) is held with its center at the point \(x=0, y=3.00 \mathrm{m} .\) (a) What are the magnitude and direction of the net gravitational force due to these objects on a third uniform sphere with mass 0.500 \(\mathrm{kg}\) placed at the point \(x=4.00 \mathrm{m}, y=0 ?(6)\) Where, other than infinitely far away, could the third sphere be placed such that the net gravitational force acting on it from the other two spheres is equal to zero?

A uniform, spherical, \(1000.0-\mathrm{kg}\) shell has a radius of 5.00 \(\mathrm{m}\) . (a) Find the gravitational force this shell exerts on a \(2.00-\mathrm{kg}\) point mass placed at the following distances from the center of the shell: (i) \(5.01 \mathrm{m},(\mathrm{ii}) 4.99 \mathrm{m},\) (iii) 2.72 \(\mathrm{m} .\) (b) Sketch a qualitative graph of the magnitude of the gravitational force this sphere exerts on a point mass \(m\) as a function of the distance \(r\) of \(m\) from the center of the sphere. Include the region from \(r=0\) to \(r \rightarrow \infty\) .

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.