mass of Earth = 5.9742 × 10^24 kilograms
= 5 974 200 000 000 000 000 000 000 kg
radius of Earth = 6 378.1 kilometers
Gravity for 1kg mass: 9.81N
Force=1000kg x gravity
Force=1000kg x gravity
Because the acceleration of gravity on the surface of any given body depends on the mass of the body and its radius ... the distance of the surface from the center. Mars' mass ... about 11% of Earth's ... and Mars' radius ... about 53% of Earth's ... combine to produce about 38% of Earth's gravitational acceleration at the surface of Mars.
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
The force of gravity is proportional to the masses of the two objects and inversely proportional to the square of the distance between them. As a result, a gravity of one fourth that on the surface of the Earth would be observable at an altitude equal to the radius of the Earth, i.e. 6400 km. Note: This is up, in the air, not down, into the Earth. This distance is in outer space.In the other direction, the force of gravity gets smaller as one goes deeper into the earth. This is because the mass outside your current radius (as you descend) does not contribute. The mass inside your current radius is proportional to the cube of the radius. Gravity is proportional to this mass divided by the square of the radius. Therefore, gravity decreases linearly with the radius. So the acceleration of gravity is equal to one fourth its value at the earth's surface at one fourth of the earth's radius, or a depth of approximately 4800 km.
Assume an initial gravity of 1 unit. Calculate the change in radius from the change in volume. That would be the third root of the volume. Divide by the radius squared.Also, multiply by the increase in mass, i.e., by a factor of 25. Since you started off by 1 unit of gravity, the final result will tell you by what factor the force of gravity increased.
This is hard to calculate precisely, due to the fact that Earth's density increases towards the center. However, you make a simplified calculation, by assuming a uniform density. Just calculate the ratio of the volume (and therefore, of mass) of a sphere which has half the radius of the Earth, and calculate the gravitational attraction (once again, you only need a ratio, compared to the complete Earth) on that object.
Because the acceleration of gravity on the surface of any given body depends on the mass of the body and its radius ... the distance of the surface from the center. Mars' mass ... about 11% of Earth's ... and Mars' radius ... about 53% of Earth's ... combine to produce about 38% of Earth's gravitational acceleration at the surface of Mars.
no
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
The radius is 3.45
because all are measured at the same radius from the earths cog, if you doubled this distance, the acceleration would be only one quarter that of the surface
The radius of Earth in kilometers is 6400 km.The earth is big in 6400 km in radius above its surface.
We can know through two ways. First, we know the moon's mass and radius. From there it is fairly simple to calculate surface gravity. Second, we have sent probes and people to the moon, making it possible to actually measure the strength of gravity.
2(pi)radius^2
The force of gravity is proportional to the masses of the two objects and inversely proportional to the square of the distance between them. As a result, a gravity of one fourth that on the surface of the Earth would be observable at an altitude equal to the radius of the Earth, i.e. 6400 km. Note: This is up, in the air, not down, into the Earth. This distance is in outer space.In the other direction, the force of gravity gets smaller as one goes deeper into the earth. This is because the mass outside your current radius (as you descend) does not contribute. The mass inside your current radius is proportional to the cube of the radius. Gravity is proportional to this mass divided by the square of the radius. Therefore, gravity decreases linearly with the radius. So the acceleration of gravity is equal to one fourth its value at the earth's surface at one fourth of the earth's radius, or a depth of approximately 4800 km.
109 Actually, no. 109 would probably be for Jupiter. For Earth, hundreds of Earth's surface could fit in the sun's radius.
-- If the mass of Mars increases, then its surface gravity also increases. -- If the mass of Mars decreases, then its surface gravity also decreases. -- So long as its radius does not change, the acceleration due to gravity on or near the planet's surface is directly proportional to its mass.
No. The gravitational pull at the surface of a planet depends on that planet's mass and radius. Jupiter has the strongest gravity of any planet in the solar system: 2.53 times the surface gravity on Earth. Mercury has the weakest surface gravity at just 37% the gravity on Earth.