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How much will be the weight of a body at the centre of the earth?
Here gₔ = 1% of g = g/100 But gₔ = g(1 - d/R) g/100 = g(1 - d/R) or d/R = 1 - 1/100 = 99/100 d = 99/100 x R = 99/100 x is 6400 = 6336 km
Here gₔ = 1% of g = g/100
See lessBut gₔ = g(1 – d/R)
g/100 = g(1 – d/R)
or d/R = 1 – 1/100 = 99/100
d = 99/100 x R = 99/100 x is 6400 = 6336 km
Express law of gravitation in vector form. What are its implications?
Vector Form of Newton's Law of Gravitation Consider two particles, A and B, with masses m₁ and m₂, separated by a distance r, as shown in the figure. In vector form, Newton's law of gravitation describes the gravitational force between these two particles. The force is attractive, meaning that partiRead more
Vector Form of Newton’s Law of Gravitation
Consider two particles, A and B, with masses m₁ and m₂, separated by a distance r, as shown in the figure.
In vector form, Newton’s law of gravitation describes the gravitational force between these two particles. The force is attractive, meaning that particle m₁ is pulled towards m₂, and vice versa.
The negative sign in the equation signifies that the force is directed opposite to the position vector, confirming that gravitational force is always attractive.
This law further implies that the gravitational forces acting between two particles form an action-reaction pair. These forces are directed along the line joining the centers of the two particles, making gravitational force a central force.
See lessWrite expressions for the gravitational force exerted by the earth on a point mass m located above, below and on the surface of the earth. Hence deduce expression for g on the earth’s surface.
Acceleration Due to Gravity of the Earth The Earth is considered a sphere with radius R and uniform density rho. Its mass is proportional to its volume and density. 1. At Points Above the Earth's Surface For a point mass m located outside the Earth at a distance r from its center, the gravitationalRead more
Acceleration Due to Gravity of the Earth
The Earth is considered a sphere with radius R and uniform density rho. Its mass is proportional to its volume and density.
1. At Points Above the Earth’s Surface
For a point mass m located outside the Earth at a distance r from its center, the gravitational force follows the shell theorem. The force acts as if the entire mass of the Earth is concentrated at its center.
2. At Points Below the Earth’s Surface
At a point inside the Earth, at depth d below the surface, the point lies outside a smaller sphere of radius r and within an outer shell. The shell exerts no force on the point, and only the inner sphere contributes to the gravitational force.
3. At Points on the Earth’s Surface
For a point mass m on the surface, the distance from the Earth’s center is equal to its radius R .
The acceleration due to gravity, denoted by g , is defined as the acceleration experienced by a mass under the influence of Earth’s gravity. It depends on the Earth’s mass and radius.
See lessObtain an expression for the acceleration due to gravity g in terms of mass of the earth M and gravitational constant G.
Acceleration Due to Gravity on the Earth's Surface The Earth is assumed to be a sphere with mass M and radius R . A body of mass m placed on the surface of the Earth experiences a gravitational force of attraction due to the Earth. According to the law of gravitation, this force acts as if the entirRead more
Acceleration Due to Gravity on the Earth’s Surface
The Earth is assumed to be a sphere with mass M and radius R . A body of mass m placed on the surface of the Earth experiences a gravitational force of attraction due to the Earth.
According to the law of gravitation, this force acts as if the entire mass of the Earth is concentrated at its center, as per the shell theorem. The gravitational force generates an acceleration g in the body, known as the acceleration due to gravity.
Using Newton’s second law of motion, the acceleration g depends only on the Earth’s mass and radius and is independent of the mass, size, or shape of the body falling under gravity.
See lessExplain how the mass and average density of the earth can be estimated from the knowledge of G?
Mass of the Earth: The mass of the Earth can be calculated using the relationship between gravitational acceleration g, the gravitational constant G , and the Earth's radius R . By substituting the known values of g, G, and R, the Earth's mass is determined to be approximately 6 \times 10²⁴ kg. ThisRead more
Mass of the Earth:
The mass of the Earth can be calculated using the relationship between gravitational acceleration g, the gravitational constant G , and the Earth’s radius R . By substituting the known values of g, G, and R, the Earth’s mass is determined to be approximately 6 \times 10²⁴ kg.
This calculation was made possible when the value of G was first experimentally determined by Cavendish, earning him the distinction of being the first person to weigh the Earth.
Average Density of the Earth:
The average density of the Earth can also be estimated using its mass and volume. Although the density of the Earth’s upper layers is around 2700 kg/m³, the inner layers have much higher densities, leading to an overall average density of about 5500 kg/m³.
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