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Gravity is greater at the what?
Gravity increases from about 9.780 m/s2 at the Equator to about 9.832 m/s2 at the poles. This means an object will weigh about 0.5% more at the poles than at the Equator.
Why does the weight of an object change when it is moved from the equator to poles?
say mass(m) = 100 kgvelocity(v) at equator = 464.6 metres / secondradius(r) to earth surface = 6 371 000 metresacceleration due to gravity (g) = 9.82 (m / s) / s.the force of attraction (f) anywhere on earths surface, = m * g = 100 * 9.82 = 982 newtons.the force of repulsion / centripetal force (f) at the equator = mass * (v^2) / r =3.39 newtons
Is Kuwait near to the equator?
No. Saudia Arabia is north of the equator.
What is 90 percent of Earths gravity?
8,9 m/s2 The gravitational attraction on the surface of Venus.
How many times can Saturn fit into the sun?
The Sun has a volume of 1.4122×10^27 m³ (1,300,000 Earths) Saturn has a volume of 8.2713×10^14 km³ (763.59 Earths) So, approximately 1,702 Saturns could fit inside the Sun if it were a hollow space.The Sun has a volume of 1.4122×10^27 m³ (1,300,000 Earths)Saturn has a volume of 8.2713×10^14 km³ (763.59 Earths)So, approximately 1,702 Saturns could fit inside the Sun if it were a hollow space.
What fraction of the Earth's surface is 1 meter?
A meter is a measure of length, not area. The equator is approximately 40,000,000 m long, so one meter is 1 forty millionth of the equator.
What is earths satellite that starts with m?
moon
If you weighed 100 pounds on Earth you would weigh pounds on Saturn.?
106 pounds Saturn's gravity acceleration at the equator is about 10.4 m/s2. Earths gravity acceleration is 9.81 m/s2. So Saturn's gravity is about 10.4/9.81 = 1.06 times stronger than Earth's. That would put a weight of 100 lb on Earth, on 106 lb on the surface of Saturn at the equator.
Gravity is greater at the what?
Gravity increases from about 9.780 m/s2 at the Equator to about 9.832 m/s2 at the poles. This means an object will weigh about 0.5% more at the poles than at the Equator.
Standing at the equator What is the centripetal force you experience due to the rotation of the earth How much does this force compare to the gravitational force you experience would it be noticeabl?
say your mass (m) is 100 kgvelocity (v) at equator = 40 030 000 / 86164 = 464.6 metres / secearth radius(r) = 6 371 000 metresacceleration due to gravity at earths surface (g) = 9.82 (m/s)/s.gravity force at earth surface ( f = m * g) = 100 * 9.82 = 982 newtons.centripetal force at equator f = m * ((v^2)/r)f = 100 * ((464.6^2)/6 371 000)f = 3.39 newtons
What is 13CM converted into meters?
the earth at its equator 1 m=100 cm 13cm=0.13 m
What produced oxygen in earths atmosphere?
yo m omma
Why does the weight of an object change when it is moved from the equator to poles?
say mass(m) = 100 kgvelocity(v) at equator = 464.6 metres / secondradius(r) to earth surface = 6 371 000 metresacceleration due to gravity (g) = 9.82 (m / s) / s.the force of attraction (f) anywhere on earths surface, = m * g = 100 * 9.82 = 982 newtons.the force of repulsion / centripetal force (f) at the equator = mass * (v^2) / r =3.39 newtons
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
What is the G force of the Earth at the equator?
The (tangential) velocity of a point at the equator is v=465 m/s. The acceleration of gravity g is diminished by v2/R = 0,033 (m/s2), where R is the earth radius which is 6.400 km approximately.
What is the affect of rotation of the earth on acceleration due to gravity?
Gravity is pretty constant figure anywhere on earth, essentially its dependent on your distance from the center of gravity of the earth, nominally it will produce an acceleration of 9.81((m/s/)/s) at the earths surface. Gravity is dependent on mass and independent of motion , ie mass of earth attracting mass of person , attraction being proportional to total mass of both and distance between their centers of gravity. However , a very small opposing acceleration outwards is experienced because you are rotating about the earths axis (centrifugal action) , its maximum effect is on the equator and zero effect at the poles. this acceleration can be calculated from: a = (v^2)/r where: a = acceleration ((m/s)/s) v = velocity (m/s) r = radius or normal distance from earths axis (m)
Is Kuwait near to the equator?
No. Saudia Arabia is north of the equator.
