What happens to the gravitational field strength as you increase the radius?

What If a planet has twice the mass of Earth its radius would have to be larger by a factor of 2 for the gravitational field strength at the planet and surface to be the same as on Earth and s?

If a planet has twice the mass of Earth, its radius would need to be larger than Earth's to maintain the same gravitational field strength at its surface. Specifically, to achieve equivalent gravitational acceleration, the radius must increase by a factor of about 1.414 (the square root of 2), not 2. This is because gravitational field strength is directly proportional to mass and inversely proportional to the square of the radius (g = G * M / r²). Therefore, a radius larger by a factor of 2 would actually result in a lower gravitational field strength than that of Earth.

If a planet has twice the mass of Earth its radius would have to be larger by a factor of 2 for the gravitational field strength at the planet and surface to be the same as on Earth and surface?

For a planet to have the same gravitational field strength at its surface as Earth while having twice its mass, its radius must increase. The gravitational field strength ( g ) is given by the formula ( g = \frac{G \cdot M}{R^2} ), where ( G ) is the gravitational constant, ( M ) is mass, and ( R ) is radius. If the mass ( M ) is doubled, to maintain the same gravitational field strength ( g ), the radius ( R ) must be increased by a factor of ( \sqrt{2} ), not 2. Therefore, the radius would need to be larger by a factor of approximately 1.414.

What is the Gravitational strength of the sun?

You need to use the radius and the mass :P

What is the value of gravitational field strength on a planet with half Earth's mass and half its radius?

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.

What if planet has twice the mass of Earth its radius would have to be larger by a factor of 2 for the gravitational field strength at the planet and surface to be the same as on Earth and surface t?

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.

What does the gravitational field strength on a planet depend on?

The gravitational field strength on a planet depends on its mass and the distance from the planet's center. The greater the planet's mass, the stronger the gravitational field, and the closer you are to the planet's center, the stronger the gravitational field.

Why is the gravitational field strength of earth and moon are different?

The gravitational field strength of Earth and the Moon differs because each celestial body has its own mass and radius. Earth is more massive and has a larger radius compared to the Moon, leading to a stronger gravitational field on Earth. The gravitational field strength decreases with distance from the center of the body, so being closer to Earth results in a stronger gravitational pull compared to being closer to the Moon.

What is the field strength at a distance one Earth radius beyond 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.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.

As the diameter of the circle increases what happens to the circumference?

The circumference of a circle increases with an increase in the radius as it is directly proportional its radius.

Jupiter has about 300 times the mass of Earth and about ten times Earth's radius. What's the Estimated size of g on the surface of Jupiter?

Since gravitational forces between two masses are proportional to m1 & m2, a 300x increase in planetary mass would INCREASE the gravitational force on an object by the same factor: 300x compared to earth. Since gravitational forces are also proportional to 1/(radius squared), a 10x increase in planetary radius would DECREASE the gravitational force by a factor of 100x (10 squared), at the planet's surface. So an object on such a planet would experience gravitational forces 3x greater than those on earth. Since gravitational forces between two masses are proportional to m1 & m2, a 300x increase in planetary mass would INCREASE the gravitational force on an object by the same factor: 300x compared to earth. Since gravitational forces are also proportional to 1/(radius squared), a 10x increase in planetary radius would DECREASE the gravitational force by a factor of 100x (10 squared), at the planet's surface. So an object on such a planet would experience gravitational forces 3x greater than those on earth.

What happens to the surface temperature of a planet if you decrease its orbital radius?

If you decrease a planet's orbital radius, its surface temperature will increase.

What happens to the size of the atomic radius as you go down the noble gas family?

The atomic radius increase down, from helium to radon.