Why is the value of acceleration g in free fall smaller at the equator than at the North Pole?

The rotation of the Earth at the equator produces a large centrifugal force that cancels out part of the value of centripetal acceleration g.

The rotation of the Earth produces a centrifugal force that cancels out a portion of the value of the gravitational acceleration g.

Note that at the equator, the centrifugal force generated by the rotation of the Earth is greater than that of the North Pole because of the large radius of rotation.

According to the ** conversion in the physical gravitational force, that is, gm=gr 2, g=gm r 2 is obtained, because the earth is not a regular sphere, but the equatorial part is convex and has a larger radius, while the poles are slightly flattened and the radius is smaller, so the g value is smaller at the equator.

In fact, as long as you understand one problem, that is, gravity and the centripetal force of the earth's rotation are both components of gravitational force!

And there is no centripetal force at the poles, so when there is no centripetal force at the poles, g = gm ground r ground 2, as for everywhere else there is a centripetal force, so the value of g is smaller at the equator than the north pole, in fact, everywhere is smaller than the north pole. As for the size, it is not difficult to find it using the parallelogram rule. In addition, although the earth is elliptical, the difference between the long and short semi-axes is very small and can be ignored under normal circumstances.

Since g does not change much with latitude, the gravitational acceleration of an object at sea level at latitude 45° will be accurately measured internationally g = m seconds; As a standard value for acceleration due to gravity. In solving problems near the earth's surface, g is usually taken as a constant, and g = meter seconds can be taken in general calculations. Theoretical analysis and precise experiments show that the value of gravitational acceleration g increases slightly with the increase of latitude, for example, near the equator g = m second, Guangzhou g = m second.

Wuhan g = meter second.

Shanghai g = meter second.

Tokyo g = meter seconds.

Beijing g = meter second.

New York g = meter seconds.

Moscow g = meter seconds.

Arctic g = meter seconds.

The conversion relationship between the different unit systems of gravitational acceleration g is: gravitational acceleration g = ; = 981cm/s; = ;

Note: The picture shows a kind of gravitational acceleration test sheet measured.

The gravitational acceleration on the surface of the Moon is about m ssup2; , which is about one-sixth of the Earth's gravity.

<> centripetal acceleration of an object at the equator is not equal to g, its centripetal acceleration is the acceleration of the gravitational force.

g is much smaller, and the physical difference at the equator is subject to the gravitational pull of the earth.

It is equal to the sum of centripetal acceleration and gravitational acceleration multiplied by the mass void of the object, i.e., gmm r2 = m·an+mg.

The equator refers to the rotation of points on the Earth's surface with the Earth.

The resulting trajectory has the longest circumference of the circumference, with a radius of kilometers in circumference. The equator is the dividing line between the northern and southern hemispheres, which divides the earth into two hemispheres, north of the northern hemisphere and south of the southern hemisphere.

The gravitational force of the earth at the equator is completely divided into mu and perturbation into the same direction of gravity and centripetal force.

And because the autobiography of the earth is relatively slow.

So the centripetal force is small.

Then the centripetal acceleration of a respectable object at the equator.

Much less than the acceleration of gravity.

Less than, because it is farther from the center of the earth.

<> the centripetal acceleration of an object at the equator is not equal to g, its centripetal acceleration is much smaller than the acceleration g of the gravity force of the terrestrial fiber, and the gravitational pull of the object at the equator is equal to the sum of the centripetal acceleration and the gravitational acceleration multiplied by the mass of the object, i.e., gmm r2=m·an+mg.

The equator refers to the longest circumference line in the trajectory of a point on the Earth's surface with the rotation of the Earth, with a radius of kilometers in circumference. The equator is the dividing line between the northern and southern hemispheres, which divides the earth into the northern and southern hemispheres, with the northern hemisphere to the north and the southern hemisphere to the south.

First of all, you need to understand.

Centripetal acceleration.

It refers to doing. Curved motion. Target. Object.

In order to be able to "turn the corner", what is needed is pointed to that.

Curve. Center.

Imagine that if there is no force, the object is done.

Linear motion. Then, centripetal acceleration is only for the purpose of maintaining the object.

Motion. a=(v 2) r, which is determined by the velocity and .

Radius. Decided, for.

Equator. on the object, g is.

Universal gravitation. The value of the acceleration caused by subtracting the centripetal acceleration, that is, the part of the gravitational force that provides the centripetal acceleration, and the rest is what we feel.

Gravity. For the specific question, use an autobiography.

Cycle. 24 hours and can be calculated, without taking into account g

You can understand.

Gravitational force = gravity + centripetal force i.e.: gmm (r2) = mg + m (w2) r = mg + ma

This passage? If you can, it's easy to explain.

Since the gravitational force on the object is m*(a+g), to make the object float, that is, the gravitational force is all used as a centripetal force, there is m*(a+g)=1 2*m*(w2) 2*r

In the actual situation, the centripetal force is only ma=1 2*m*(w1) 2*r and compare w2 and w1 to get the answer.

gmm/（r2）=mg+m(w2)r=mg+magmm/（r2）=mr(w'2) The gravitational force is all used to provide the centripetal force w'2=gm/(r3)=(g+a)/r

w = under the root number [a r] = under the root number [gm (r3)] = under the root number [g+a] is wrong.

When the object is on the equator, part of the earth's gravitational force on the object acts as gravity, and part of it acts as a centripetal force, and the speed on the upper line of the equator is the largest, so the centripetal force is the largest at this time, and the gravitational force remains unchanged, so the gravitational force is the smallest, and the gravitational acceleration is naturally the smallest.

The so-called g refers to the average gravitational acceleration on the earth's surface, which increases from the equator to the poles.

When the object rotates with the earth, the object at the equator is subjected to gravitational and supporting force, and the supporting force is equal to the gravitational force, i.e., f-g=m 2r=m(2 n)2r The object is only subject to gravitational force when it "floats", so f=ma so a =g+a and since g+a= 2r=(2 n)2r simultaneous solution n n =g+a a so the answer is: g+a a

Because the definition of free fall is the motion of a regular object only under the action of gravity, the initial velocity is zero, and the acceleration due to gravity isg≈

When an object rotates with the Earth at the equator, it has a=

r；When an object rotates with the Earth, the object at the equator is subjected to gravitational and supporting forces, and the supporting force is equal to the gravitational force, i.e.

f-mg=ma；

When an object "floats", it is only subject to gravitational pull, therefore.

f=ma so a = g+a, that is, when the object "floats", the acceleration of the object is g+a, then there is: g+a= 22r solution:

aa+g

Calculation process:

Let the centripetal force be f, the mass be m, the rotation period of the earth be t, and the radius of the earth be rf=mr(2 t) 2

ma=mr(2π/t)^2

a=r(2π/t)^2

r≈t=≈a≈

Analysis: This centripetal acceleration is much less than the acceleration due to gravity, so people do not feel the rotation on Earth.

In this case, the gravitational force on the object completely acts as centripetal acceleration, i.e., g=a=v2 r, so the linear velocity of the earth's rotational speed is (gr) 1 2where r is the radius of the earth.

You can understand. Gravitational force = gravity + centripetal force i.e.: gmm (r2) = mg + m (w2) r = mg + ma

This passage? If you can, it's easy to explain.

Since the gravitational force experienced by the object is m*(a+g), to make the object float, that is, the gravitational force is all used as the centripetal force of the source, there is m*(a+g)=1 2*m*(w2) 2*r

In the actual situation, the centripetal force is only ma=1 2*m*(w1) 2*r and compare w2 and w1, and you can get the answer.

Acceleration due to gravity at the equator g = centripetal acceleration a = 4 2r t 2 = gmm r 2 = ma a = gm r 2 centripetal acceleration of near-earth satellites is equal to g, and centripetal acceleration of geostationary satellites a