Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A typical adult human has a mass of about 70 kg. (a) What force does a full moon exert on such a human when it is directly overhead with its center 378,000 km away? (b) Compare this force with the force exerted on the human by the earth.

Short Answer

Expert verified
(a) The moon exerts about 1.29 x 10^-4 N; (b) Earth exerts a much greater force at 686.7 N.

Step by step solution

01

Understand Gravitational Force Formula

The force exerted by a celestial body like the moon on an object is given by Newton's law of universal gravitation: \[ F = \frac{G \times m_1 \times m_2}{r^2} \]where \( F \) is the gravitational force, \( G \) is the gravitational constant \((6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2)\), \( m_1 \) is the mass of the moon, \( m_2 \) is the mass of the human, and \( r \) is the distance between the centers of the two masses.
02

Input Known Values for Moon-Human Interaction

For this calculation, use:- \( m_1 = 7.342 \times 10^{22} \, \text{kg} \) (mass of the moon),- \( m_2 = 70 \, \text{kg} \) (mass of the human),- \( r = 378,000 \, \text{km} = 3.78 \times 10^8 \, \text{m} \).Insert these values into the gravitation equation to find \( F \) exerted by the moon.
03

Calculate the Force Exerted by the Moon

Using the formula:\[ F = \frac{6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \times 7.342 \times 10^{22} \, \text{kg} \times 70 \, \text{kg}}{(3.78 \times 10^8 \, \text{m})^2} \]\[ F = 1.29 \times 10^{-4} \, \text{N} \].Thus, the force exerted by the full moon on the human is approximately \( 1.29 \times 10^{-4} \, \text{N} \).
04

Calculate the Force Exerted by the Earth

Now, use Earth's gravitational force using:\[ F = m_2 \times g \]where \( g = 9.81 \, \text{m/s}^2 \) is Earth's gravitational acceleration.\[ F = 70 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 686.7 \, \text{N} \].The gravitational force exerted by the Earth on the human is approximately \( 686.7 \, \text{N} \).
05

Compare the Two Forces

Compare the gravitational force exerted by the moon with that by the Earth.The force from the moon \((1.29 \times 10^{-4} \, \text{N})\) is much smaller than the force from the Earth \((686.7 \, \text{N})\).\[ \frac{F_{\text{moon}}}{F_{\text{earth}}} = \frac{1.29 \times 10^{-4}}{686.7} \approx 1.88 \times 10^{-7} \].The force due to the moon is negligible compared to the force due to Earth.

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 Vaia!

Key Concepts

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

Newton's Law of Universal Gravitation
Understanding the basics of gravitational attraction is essential in physics. Newton's law of universal gravitation describes how all objects in the universe attract each other with a force. This force is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

The formula is expressed as: \[ F = \frac{G \times m_1 \times m_2}{r^2} \]where:
  • \( F \) is the gravitational force between the objects
  • \( G \) is the gravitational constant, \( 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \)
  • \( m_1 \) is the mass of the first object
  • \( m_2 \) is the mass of the second object
  • \( r \) is the distance between the centers of the two objects
This law applies universally, from the attraction between celestial bodies like Earth and the moon, to the forces between objects on Earth itself.
Mass
Mass is a measure of the amount of matter in an object and it remains constant regardless of its location in the universe. It differs from weight, which is the gravitational force experienced by an object in a gravitational field.
- **Constant Property**: Mass does not change with the object's environment, unlike weight which can vary depending on gravitational pull. - **Units**: The standard unit for mass in the metric system is the kilogram (kg). - **Impact on Gravitational Force**: Greater mass means a greater gravitational force exerted, according to Newton's law. More massive objects will exert and experience stronger gravitational attractions.
In the universe, mass serves as one of the key quantities determining the gravitational interaction between objects. For example, the mass of a human (such as 70 kg) is critical in calculating the gravitational force exerted by larger celestial bodies like the moon.
Gravitational Constant
The gravitational constant \( G \) is a fundamental constant that plays a crucial role in calculating gravitational force between two masses. It helps establish the relationship between mass, force, and distance in Newton's law of universal gravitation.
- **Value**: \( G = 6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \) is considered one of the smallest constants in physics. It indicates how weak the gravitational force is compared to other forces like electromagnetism.- **Importance**: Without \( G \), we wouldn't be able to quantify the gravitational pull between objects the way we do.- **Historical Aspect**: Henry Cavendish first measured \( G \) in the 18th century through his famous torsion balance experiment, thereby allowing for the calculation of Earth's mass.
The gravitational constant is pivotal for understanding the forces that keep planets in orbit and govern the dynamics of galaxies, stars, and other celestial phenomena.
Moon-Earth Comparison
When we compare the gravitational force exerted by the moon and Earth on a human, we see a striking difference. The force exerted on a person standing on the Earth's surface is vastly greater than the force exerted by the moon.
- **Moon's Force**: Even when the moon is directly overhead, its gravitational pull on a human is relatively minuscule. For a 70 kg person, it's around \( 1.29 \times 10^{-4} \, \text{N} \).- **Earth's Force**: The gravitational force earth exerts on the same person is approximately \( 686.7 \, \text{N} \). This is the force we feel as weight due to Earth's gravitational acceleration, \( g = 9.81 \, \text{m/s}^2 \).- **Ratio**: The comparison shows that the moon's gravitational influence on a human is almost negligible. The ratio, \( \frac{F_{\text{moon}}}{F_{\text{earth}}} \approx 1.88 \times 10^{-7} \), illustrates just how dominant Earth's gravitational pull is.
Such comparisons highlight the overwhelming influence Earth's mass has on gravitational interactions experienced by objects on its surface.

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

You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surface, the ship's orbital speed is 4900 m/s. By observing the planet, you determine its radius to be 4.48 \(\times\) 10\(^6\) m. You then land on the surface and, at a place where the ground is level, launch a small projectile with initial speed 12.6 m/s at an angle of 30.8\(^\circ\) above the horizontal. If resistance due to the planet's atmosphere is negligible, what is the horizontal range of the projectile?

Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface. Model a spherically symmetric planet, with the same radius as the earth, as having a density that decreases linearly with distance from the center. Let the density be 15.0 \(\times\) 10\(^3\) kg/m\(^3\) at the center and 2.0 \(\times\) 10\(^3\) kg/m\(^3\) at the surface. What is the acceleration due to gravity at the surface of this planet?

There is strong evidence that Europa, a satellite of Jupiter, has a liquid ocean beneath its icy surface. Many scientists think we should land a vehicle there to search for life. Before launching it, we would want to test such a lander under the gravity conditions at the surface of Europa. One way to do this is to put the lander at the end of a rotating arm in an orbiting earth satellite. If the arm is 4.25 m long and pivots about one end, at what angular speed (in rpm) should it spin so that the acceleration of the lander is the same as the acceleration due to gravity at the surface of Europa? The mass of Europa is 4.80 \(\times\) 10\(^{22}\) kg and its diameter is 3120 km.

Your starship, the \(Aimless\) \(Wanderer\), lands on the mysterious planet Mongo. As chief scientist-engineer, you make the following measurements: A 2.50-kg stone thrown upward from the ground at 12.0 m/s returns to the ground in 4.80 s; the circumference of Mongo at the equator is 2.00 \(\times\) 10\(^5\) km; and there is no appreciable atmosphere on Mongo. The starship commander, Captain Confusion, asks for the following information: (a) What is the mass of Mongo? (b) If the \(Aimless\) \(Wanderer\) goes into a circular orbit 30,000 km above the surface of Mongo, how many hours will it take the ship to complete one orbit?

A rocket with mass 5.00 \(\times\) 10\(^3\) kg is in a circular orbit of radius 7.20 \(\times\) 10\(^6\) m around the earth. The rocket's engines fire for a period of time to increase that radius to 8.80 \(\times\) 10\(^6\) m, with the orbit again circular. (a) What is the change in the rocket's kinetic energy? Does the kinetic energy increase or decrease? (b) What is the change in the rocket's gravitational potential energy? Does the potential energy increase or decrease? (c) How much work is done by the rocket engines in changing the orbital radius?

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.

Sign-up for free