The gravitational field strength at the Earth's surface is approximately 10 N/kg (or 10 m/s^2) due to the mass of the Earth and the distance from its center. This value is a result of the acceleration experienced by objects in free fall near the Earth's surface.
I assume you mean, of the gravitational field? The gravitational field is inversely proportional to the square of the distance. At a distance of 1 Earth radius, the distance from the center of the Earth is twice the distance at the Earth's surface; thus, the field strength is 1/4 what it is on the surface. If at the surface the field strength is about 9.8 meters per second square, divide that by 4 to get the field strength at a distance of one Earth radius from the surface.I assume you mean, of the gravitational field? The gravitational field is inversely proportional to the square of the distance. At a distance of 1 Earth radius, the distance from the center of the Earth is twice the distance at the Earth's surface; thus, the field strength is 1/4 what it is on the surface. If at the surface the field strength is about 9.8 meters per second square, divide that by 4 to get the field strength at a distance of one Earth radius from the surface.I assume you mean, of the gravitational field? The gravitational field is inversely proportional to the square of the distance. At a distance of 1 Earth radius, the distance from the center of the Earth is twice the distance at the Earth's surface; thus, the field strength is 1/4 what it is on the surface. If at the surface the field strength is about 9.8 meters per second square, divide that by 4 to get the field strength at a distance of one Earth radius from the surface.I assume you mean, of the gravitational field? The gravitational field is inversely proportional to the square of the distance. At a distance of 1 Earth radius, the distance from the center of the Earth is twice the distance at the Earth's surface; thus, the field strength is 1/4 what it is on the surface. If at the surface the field strength is about 9.8 meters per second square, divide that by 4 to get the field strength at a distance of one Earth radius from the surface.
The gravitational field strength on Mercury is approximately 3.7 m/s^2. This means that objects on the surface of Mercury experience a gravitational force that is 3.7 times that of Earth's gravitational force.
The gravitational field strength at the surface of Earth is approximately 9.81 m/s^2, which is effectively the same as the acceleration due to gravity or the acceleration of free fall. This value is commonly used to represent the rate at which objects accelerate towards the Earth when dropped.
The gravitational field strength on Earth is approximately 9.81 meters per second squared (m/s^2). This value represents the acceleration due to gravity experienced by objects near Earth's surface.
The gravitational field strength of the Sun is approximately 274 m/s^2 at its surface. This means that objects near the surface of the Sun experience a gravitational acceleration of about 274 m/s^2.
Mercury's gravitational field strength is approximately 3.7 m/s^2, which is about 38% of Earth's gravitational field strength. This means that objects on the surface of Mercury would weigh less compared to Earth due to the lower gravitational pull.
I assume you mean, of the gravitational field? The gravitational field is inversely proportional to the square of the distance. At a distance of 1 Earth radius, the distance from the center of the Earth is twice the distance at the Earth's surface; thus, the field strength is 1/4 what it is on the surface. If at the surface the field strength is about 9.8 meters per second square, divide that by 4 to get the field strength at a distance of one Earth radius from the surface.I assume you mean, of the gravitational field? The gravitational field is inversely proportional to the square of the distance. At a distance of 1 Earth radius, the distance from the center of the Earth is twice the distance at the Earth's surface; thus, the field strength is 1/4 what it is on the surface. If at the surface the field strength is about 9.8 meters per second square, divide that by 4 to get the field strength at a distance of one Earth radius from the surface.I assume you mean, of the gravitational field? The gravitational field is inversely proportional to the square of the distance. At a distance of 1 Earth radius, the distance from the center of the Earth is twice the distance at the Earth's surface; thus, the field strength is 1/4 what it is on the surface. If at the surface the field strength is about 9.8 meters per second square, divide that by 4 to get the field strength at a distance of one Earth radius from the surface.I assume you mean, of the gravitational field? The gravitational field is inversely proportional to the square of the distance. At a distance of 1 Earth radius, the distance from the center of the Earth is twice the distance at the Earth's surface; thus, the field strength is 1/4 what it is on the surface. If at the surface the field strength is about 9.8 meters per second square, divide that by 4 to get the field strength at a distance of one Earth radius from the surface.
The gravitational field strength of a planet multiplied by an objects mass gives us the weight of that object, and that the gravitational field strength, g of Earth is equal to the acceleration of free fall at its surface, 9.81ms − 2.
The gravitational field strength on Mercury is approximately 3.7 m/s^2. This means that objects on the surface of Mercury experience a gravitational force that is 3.7 times that of Earth's gravitational force.
The gravitational field strength on Venus is about 8.87 m/s^2, which is about 91% of Earth's gravitational field strength. This means that objects on Venus would feel slightly lighter compared to on Earth.
If a planet has twice the mass of Earth and its radius is increased by a factor of 2, the gravitational field strength at its surface can be calculated using the formula ( g = \frac{GM}{R^2} ). Here, ( G ) is the gravitational constant, ( M ) is the mass, and ( R ) is the radius. By doubling the radius while doubling the mass, the gravitational field strength becomes ( g' = \frac{2G(2M_E)}{(2R_E)^2} = \frac{G M_E}{R_E^2} ), which equals Earth's gravitational field strength. Thus, the conditions for gravitational strength to be the same as on Earth are satisfied.
The gravitational field strength at the surface of Earth is approximately 9.81 m/s^2, which is effectively the same as the acceleration due to gravity or the acceleration of free fall. This value is commonly used to represent the rate at which objects accelerate towards the Earth when dropped.
The gravitational field strength of the Moon is about 1.6 N/kg, which is about 1/6th of the gravitational field strength on Earth.
The acceleration due to gravity on the surface Venus is 8.9m/s2 That's about 90% of Earth's.
The Earth's gravitational field pulls objects towards its center, creating the force of gravity that keeps everything on the surface of the Earth and governs the motion of celestial bodies in space. The strength of gravity decreases with distance from the Earth's surface according to the inverse square law.
The value of the gravitational field strength on a planet with half the mass and half the radius of Earth would be the same as Earth's gravitational field strength. This is because the gravitational field strength depends only on the mass of the planet and the distance from the center, not on the size or density of the planet.
The gravitational field strength of Io, one of the moons of Jupiter, is approximately 1.796 m/s^2. This value is about 1/6th of Earth's gravitational field strength.