The question asks about Newton's Law of Universal Gravitation. To answer this, we need to recall the mathematical formulation of this law, which describes the gravitational force between two objects.
Correct Option: B) The product of the masses and inversely proportional to the square of the distance between them
This option perfectly aligns with the mathematical formulation of Newton's Law of Universal Gravitation, \(F \propto \frac{m_1 m_2}{r^2}\).
To determine how the gravitational force changes when the distance between two bodies is halved, we need to recall Newton's Law of Universal Gravitation. This law describes the relationship between gravitational force, the masses of the bodies, and the distance between their centers.
Correct Option: A) It becomes four times larger
The question asks about the factors influencing the gravitational force between two masses. This can be directly answered by recalling Newton's Law of Universal Gravitation.
Newton's Law of Universal Gravitation states that every particle attracts every other particle 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.
Mathematically, this law is expressed as:
\[ F = G \frac{m_1 m_2}{r^2} \]Where:
From this formula, it is clear that the gravitational force \(F\) depends on the product of the masses (\(m_1 m_2\)) and is inversely proportional to the square of the distance (\(r^2\)) between them. The gravitational constant \(G\) is a universal constant and does not depend on the specific masses or distance.
Therefore, the gravitational force depends on their masses and the distance between them.
Correct Option: C) Their masses and the distance between them
A) Their masses, distance, and speed — Speed is not a factor in determining the gravitational force according to Newton's Law of Universal Gravitation. Gravitational force is a static property between masses at a given distance, not dependent on their motion.
B) Their masses only — While masses are a factor, the distance between them is also crucial, as the force decreases with the square of the distance.
D) The distance between them only — While distance is a factor, the magnitudes of the masses are also crucial, as the force is directly proportional to the product of the masses.
To determine what provides the centripetal force for a satellite in Earth orbit, we need to understand the fundamental forces acting on objects in space and the definition of centripetal force. Centripetal force is any force that causes an object to follow a curved path.
Correct Option: D) Gravitational attraction between the satellite and Earth provides the necessary centripetal force to keep a satellite in a stable orbit. According to Newton's Law of Universal Gravitation, every particle attracts every other particle 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. This gravitational force acts as the centripetal force, pulling the satellite towards the Earth's center and maintaining its curved path.
The weight of an object is determined by the gravitational force exerted on it. According to Newton's Law of Universal Gravitation, this force depends on the masses of the two objects and the square of the distance between their centers. We will analyze how this force changes with increasing distance.
A) It decreases. As explained, the gravitational force, and thus the weight, is inversely proportional to the square of the distance from the center of the Earth. Therefore, as the distance increases, the weight decreases.
The escape velocity is the minimum speed an object needs to escape the gravitational pull of a celestial body, like Earth, and move into space without falling back. This value is a fundamental constant for a given planet and can be calculated using a specific formula involving the gravitational constant, the mass of the planet, and its radius.
B) 112 km/s — This is the widely accepted and calculated value for the escape velocity from Earth's surface, approximately \(11.2 \text{ km/s}\).