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What would be its relative strength at the new surface if Earth's radius shrank to 0.65 its size?
Wiki User
∙ 11y ago
If you are asking about gravity, its strength depends on the mass and the distance from the centre, so the force varies as M/r^2.
If you assume the mass stays the same, gravity at the surface would be 1/(0.65^2) times stronger for that reduction in radius, so a little more than twice as strong.
Wiki User
∙ 11y ago
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How many times bigger than the earths radius is the suns radius?
the sun's radius is and half a million bigger than the radius of the sun.
When you are standing at sea level which are we closest to the earths core or moon surface?
The Earth's radius is only 4000 miles (approx.). The moon is approx. 250,000 miles away. The moon loses by a lot.
How does the earths radius affect the gravitational force on a object at its surface?
The Earth and the object exert a gravitational force on each other, but only the Earth's is big enough to measure. So, the formula for gravitational force include the distance from one body's surface to its center and the same for the other body. The length of the radius is directly proportional to the body's gravitational force.
How many earth can Jupiter accommodate if the radius of Jupiter is 11 times the radius of the earth?
Approx 1000 earths would fit into Jupiter
What does the gravitational field strength on a planet depend on?
It can be calculated on the basis of the planet's mass and its radius.
Is the earths radius and surface area increasing?
no
How big is the earths kilometer above its surface?
The radius of Earth in kilometers is 6400 km.The earth is big in 6400 km in radius above its surface.
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.
How many times is the suns radius bigger than earths surface?
109 Actually, no. 109 would probably be for Jupiter. For Earth, hundreds of Earth's surface could fit in the sun's radius.
What is the gravitational force on any object near the earths surface?
Force (newtons) = mass (kg) * acceleration (m/s/s) > Acceleration at earths surface radius = 9.82 m/s/s
How large is the second smallest planet?
MarsIts equatorial radius is 3,396.2 ± 0.1 km or 0.533 Earths,Its polar radius is 3,376.2 ± 0.1 km or 0.531 Earths,Its surface area is 144,798,500 km² or 0.284 Earths,Its volume is 1.6318 × 1011 km³ or 0.151 Earths,Its mass is 6.4185 × 1023 kg or 0.107 Earths.
The planet Neptune's size compared to the other planets?
Mercury's radius (at the equator) = 2440km or 0.3825 x Earths Venus' radius = 6052km or 0.9488 x Earths Earth's radius = 6378km Mars' radius = 3397km or 0.5323 x Earths Jupiter's radius = 71,492km or 11.21 x Earths Saturn's radius = 60,267km or 9.45 x Earths Uranus' radius = 25,557km or 4.01 x Earths Neptune's radius = 24,766km or 3.88 x Earths
What is the distance from the Earths center to a point outside the Earth where the gravitational acceleration due to the Earth is 1 26 of its value at the Earths surface?
That would have to be at a radius that is sqrt(26) = 5.1 times the Earth's physical radius, or about 32,486 kilometers (20,186 miles) from the center.
What is the distance from earths surface to the center of earth?
That's the Earth's radius. It's about 6378 kilometers at the equator. That's 3963 miles.
What is earths gravitational acceleration?
Standing at surface radius its = 9.82 (m/s)/sbut double the radius and the acceleration drops to 9.82 / ((2 / 1)2) = 2.455 (m/s)/s
How do mercury radius and mass compare with the earths?
Mercurys mean radius is around 2,439.7 km, which is around 38.39% of earth radius or 0.3829 Earths. Mercurys mass is around 3.3022×10 to the 23 kg, around 5.5% of earths mass or 0.055 Earths.
What is the acceleration due to gravity 9000000 meters away from the center of the earth?
If you have a known rate of acceleration and radius (such as at the earths surface), you can use the following equation to calculate the acceleration at another radius.a = k / ((d / r)^2)key:a = new acceleration rate ((m/s)/s)d = new radius (metres)k = known acceleration rate ((m/s)/s)r = known radius (metres)so if:d = 9 000 000 metresk = 9.82 (m/s)/s (acceleration at earths surface)r = 6 371 000 metres (radius at earths surface)then:a = 4.92 (m/s)/s
