Skip to main content
Ch 13: Newton's Theory of Gravity
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoKnight Calc 5th EditionCh 13: Newton's Theory of GravityProblema 39
Capitolo 13, Problema 39

Two spherical objects have a combined mass of 150 kg. The gravitational attraction between them is 8.00 x 10-6 N when their centers are 20 cm apart. What is the mass of each?

Guida verificata passo dopo passo
1
Start by recalling Newton's law of universal gravitation: F = \(\frac{G m_1 m_2}{r^2}\), where F is the gravitational force, G is the gravitational constant (6.674 \(\times\) 10^{-11} \ \(\text{N·m}\)^2/\(\text{kg}\)^2), m_1 and m_2 are the masses of the two objects, and r is the distance between their centers.
Substitute the known values into the formula: 8.00 \(\times\) 10^{-6} = \(\frac{(6.674 \times 10^{-11}\)) m_1 m_2}{(0.20)^2}. Note that the distance r is converted to meters (20 cm = 0.20 m).
Simplify the equation to isolate the product of the masses: m_1 m_2 = \(\frac{(8.00 \times 10^{-6}\)) (0.20)^2}{6.674 \(\times\) 10^{-11}}. This gives the value of m_1 m_2.
Use the fact that the combined mass of the two objects is 150 kg: m_1 + m_2 = 150. This provides a second equation to solve for m_1 and m_2.
Solve the system of equations: m_1 m_2 = \(\text{value from step 3}\) and m_1 + m_2 = 150. Use substitution or the quadratic formula to find the individual masses m_1 and m_2.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.
Durata del video:
8m

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Newton's Law of Universal Gravitation

This law states that every point mass attracts every other point mass in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. The formula is F = G(m1*m2)/r², where F is the gravitational force, G is the gravitational constant, m1 and m2 are the masses, and r is the distance between the centers.
Video consigliato:
Percorso guidato
04:05
Universal Law of Gravitation

Mass and Weight

Mass is a measure of the amount of matter in an object, typically measured in kilograms. Weight, on the other hand, is the force exerted by gravity on that mass. In this problem, the combined mass of the two objects is given, which is essential for determining the individual masses using the gravitational force between them.
Video consigliato:
Percorso guidato
10:19
Torque Due to Weight

Algebraic Manipulation

Algebraic manipulation involves rearranging equations to isolate variables and solve for unknowns. In this context, it is necessary to express the masses of the two objects in terms of their combined mass and the gravitational force, allowing for the calculation of each mass using the provided information.
Video consigliato:
Percorso guidato
4:06
Simplifying Algebraic Expression
Pratica correlata
Domanda del libro di testo

What is the total gravitational potential energy of the three masses in FIGURE P13.35?

1886
views
Domanda del libro di testo

Suppose that on earth you can jump straight up a distance of 45 cm. Asteroids are made of material with mass density 2800 kg/m³ . What is the maximum diameter of a spherical asteroid from which you could escape by jumping?

367
views
Domanda del libro di testo

What is the net gravitational force on the 20.0 kg mass in FIGURE P13.36? Give your answer using unit vectors.

1991
views
Domanda del libro di testo

Two spherical objects have a combined mass of 400 kg. The gravitational attraction between them is 6.00 x 10-7 N and their gravitational potential energy is ₋1.20 x 10-6 J. What is the mass of each?

306
views
Domanda del libro di testo

A rogue band of colonists on the moon declares war and prepares to use a catapult to launch large boulders at the earth. Assume that the boulders are launched from the point on the moon nearest the earth. For this problem you can ignore the rotation of the two bodies and the orbiting of the moon. What is the minimum speed with which a boulder must be launched to reach the earth? Hint: The minimum speed is not the escape speed. You need to analyze a three-body system.

1298
views
Domanda del libro di testo

What is the total gravitational potential energy of the three masses in FIGURE P13.36?

1116
views