The weight of an object is slightly less at the equator than at the poles because of the earth's tilt on its axis.
The earth is not a perfect sphere. The spin of the earth causes it to buldge out at the equator, which means the equator is further from the center of the earth then the poles are. The further an object is from the center of mass of another object, the less effect the gravity of those objects will have on each other. So at the equator, an object is being effected less by the gravity of the earth then it is at the 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
It will be approx 98.1 Newtons. The weight will vary noticablyover the surface of the earth: more at the poles, less near the equator,with altitude: less at higher altitudes,with surrounding geography and weather: eg more after heavy rain.
I would say that buying sugar at the equator is more profitable as gravitational pull is less at the equator (you can say almost zero) as compared to the poles (where the gravitational pull is the highest) , so sugar will weigh more at the poles than at the equator. Therefore, we can say that the price of sugar will be more at the poles than at the equator. The sugar will WEIGH less at the equator , so the price will also be less there. Hence, buying sugar at the equator is more profitable.NOTE: SUGAR WILL HAVE SAME MASS AT THE EQUATOR AS WELL AS AT THE POLES.====================================Beautiful. But, since the sell-price of anything reflects both the cost of producing it AND the cost of transporting it to market, have you considered the cost of shipping sugar to the equator from where it grows, compared to the cost of shipping sugar to the poles from where it grows ? When I worked briefly in an industrial complex in northern Alaska ten years ago, the price of a gallon of gasoline there was already over $6 .By the way ... before I go ... the force of gravity at the poles is not "almost zero", and the apparent weight of an object at the pole compared to its weight at the equator is greater on only the order of 1 percent, on account of both the greater radius and the effect of Earth's rotation at the equator.
Yes. The mass of an object will stay the same, regardless of the gravity that is effecting it. But the weight of an object depends on the apparent gravity. At the poles you would weigh more than at the equator due to the earths spin. At the equator you might weigh up to 0.3% less than atthe poles. Other factors effect the local gravity such as the density of the rock beneath the person, more dense rock will give a higher gravitational field. The height above the surface will also reduce the apparent gravity.
much less
Because of centripetal acceleration you will weigh a tiny amount less at the equator than at the poles.
less gravity pull farther away from central pole
The earth is not a perfect sphere. The spin of the earth causes it to buldge out at the equator, which means the equator is further from the center of the earth then the poles are. The further an object is from the center of mass of another object, the less effect the gravity of those objects will have on each other. So at the equator, an object is being effected less by the gravity of the earth then it is at the poles.
Due to the centrifugal force caused by Earth's rotation opposing gravity for objects on the equator, objects there weigh about 0.5% less than they do on the poles. So an object that weighs 200 N at the poles weighs about 199 N on the equator.
Rotation.
greater
Well the above question is not true, The same object will weigh less at the equator than at the poles (of Earth). The force is the force of gravity and the effect is because the object placed at the poles will be nearer the center of the Earth than at the equator because the Earth is an Oblate Spheroid.
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
It will be approx 98.1 Newtons. The weight will vary noticablyover the surface of the earth: more at the poles, less near the equator,with altitude: less at higher altitudes,with surrounding geography and weather: eg more after heavy rain.
no, we do not weigh the same at the poles because as the earth is not perfectly round and is like an orange, the distance between the object and the earth's core is less due to which we weigh more on poles
no, but the electromagnetic field of the earth does.