# How fast should the Earth spin for centripital accel. to equal gravity?

• iamnotsmart
In summary, the conversation discusses the Earth's rotation, its spherical shape, and its mass in relation to the gravitational acceleration on its surface. The conversation also explores the centripetal acceleration of a person at the equator, the speed at which the Earth needs to turn for the centripetal acceleration to equal gravity, and the centripetal acceleration at different latitudes. It also references a homework problem and provides a link to a figure.
iamnotsmart

## Homework Statement

1. Homework Statement The Earth turns once around its axis in 24.0 hours. We will assume that it is perfectly spherical, with radius 6400 km. The mass of the Earth is taken to be 6.00 × 1024 kg. The gravitational acceleration on the surface is taken to be g = 9.80 m/s2 .

a) What is the centripetal acceleration of a person at the equator? How big a fraction of the gravitational acceleration does this correspond to? Remember to draw a sketch of the situation.

b) How fast would the Earth have to turn for the centripetal acceleration to be exactly equal to gravity? Give the answer in revolutions per day. Remember to draw a sketch of the situation.

c) Define θ to be the latitude, so that θ = 0 corresponds to the equator, and θ = 90◦ is the North Pole. What is the centripetal acceleration of a person standing on the surface at a given value of θ? Remember to draw a sketch of the situation.

d) Taking into account that the centripetal acceleration is towards the rotation axis, and gravity is towards the centre of the Earth, how fast should the Earth now turn for gravity to be just enough to provide the necessary centripetal acceleration (as function of θ)? Remember to draw a sketch of the situation.

e) Using the force of gravity |Fg| = G (Mm)/r^2 , where M is the mass of the Earth, m the mass of the orbiting object. Find the required speed vorbit, as a function of r, so that the object performs uniform circular motion around the Earth, under the influence of gravity. How long does it take to go round once in an orbit 900 km above ground? Remember to draw a sketch of the situation.

Reference https://www.physicsforums.com/threa...e-to-provide-centripital-acceleration.833369/

a=v^2/r
v=2*pi*r/T

## The Attempt at a Solution

So a), b), c) and e) have been solved. The biggest problem here is d). I thought you can put centripetal acceleration=g*cos theta cause it is one of the components of g. But when I put it like that, the theta goes away. But the question is how fast does the Earth move so gravity provides just enough centripetal acceleration? However the cos(theta) cancels out in the equation? I am so confused

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As you have already noticed, the pull of gravity (G) is composed of two components: one towards the equator (sin(θ)G), one towards the axis of rotation (cos(θ)G). And only the cosine component needs to be accounted for.

The radius from the axis of rotation is cos(θ)R. I'm not sure whether that causes the cosine term to cancel, but that is where you have to start. If the cosine term does cancel out,, your answer will be the same for the value as it is on the equator. Otherwise, it will be a function of theta.

Not that this should affect your response to this homework assignment, but I also notice that the problem misstates the rate of the Earth's rotation. The Earth rotates 360 degrees in a little less than one day. More specifically, it rotates once per sidereal day, about 23.934 hours.

.Scott said:
As you have already noticed, the pull of gravity (G) is composed of two components: one towards the equator (sin(θ)G), one towards the axis of rotation (cos(θ)G). And only the cosine component needs to be accounted for.

The radius from the axis of rotation is cos(θ)R. I'm not sure whether that causes the cosine term to cancel, but that is where you have to start. If the cosine term does cancel out,, your answer will be the same for the value as it is on the equator. Otherwise, it will be a function of theta.

Not that this should affect your response to this homework assignment, but I also notice that the problem misstates the rate of the Earth's rotation. The Earth rotates 360 degrees in a little less than one day. More specifically, it rotates once per sidereal day, about 23.934 hours.

My thoughts in the beginning were putting the cosine component of gravity quals to the centripetal acceleration. The centripetal acceleration is just v^2/r, but I know that the radius from the axis of rotation is R*cos(theta), which gives v=4(pi^2)*(r^2)*(cos(theta)^2)/time. This leads to a=4(pi^2)*cos(theta)/time^2. I put this equals to g*cos(theta) cause they ask about how fast the Earth has to turn so gravity provides the necessarily centripetal acceleration (as a function of theta). But as you see, cos(theta) cancels out, and I have the same equation as b) which is the same value as it is on the equator. But the task asks for a function of theta though, so I am so confused...

.Scott said:
As you have already noticed, the pull of gravity (G) is composed of two components: one towards the equator (sin(θ)G), one towards the axis of rotation (cos(θ)G). And only the cosine component needs to be accounted for.

The radius from the axis of rotation is cos(θ)R. I'm not sure whether that causes the cosine term to cancel, but that is where you have to start. If the cosine term does cancel out,, your answer will be the same for the value as it is on the equator. Otherwise, it will be a function of theta.

Not that this should affect your response to this homework assignment, but I also notice that the problem misstates the rate of the Earth's rotation. The Earth rotates 360 degrees in a little less than one day. More specifically, it rotates once per sidereal day, about 23.934 hours.

Might as well give you the figure of the whole task http://imgur.com/a/HWJ6w

iamnotsmart said:
and I have the same equation as b) which is the same value as it is on the equator
Question b) is the one I have trouble with. It doesn't specify the equator. As vectors, they cannot be equal anywhere else.
For d) there is no reason to reject your answer just because theta disappeared. Saying "as a function of theta" just tells you to allow for different values of theta, not assume this is at the equator. A constant function is still a function.

iamnotsmart said:
Might as well give you the figure of the whole task
haruspex said:
Question b) is the one I have trouble with. It doesn't specify the equator. As vectors, they cannot be equal anywhere else.
For d) there is no reason to reject your answer just because theta disappeared. Saying "as a function of theta" just tells you to allow for different values of theta, not assume this is at the equator. A constant function is still a function.

In b) I get 17 revolutions per day, which is fine. But for d) I get a function without cos(theta) when the question asks for it. I don't know if it is a trick question just to make you confused or if you really have to find a function with theta...

iamnotsmart said:
In b) I get 17 revolutions per day, which is fine.
But only if you assume it means at the equator. It should state that.
iamnotsmart said:
for d) I get a function without cos(theta)
As I wrote, don't assume theta has to appear in the answer. If you know y is a function of x, but it turns out that it is a constant, that is still, technically, a function of x, though you might say a degenerate one.
I don't think it is intended as a trick. As I wrote, they wanted you to consider all possible values of theta in d, not just the equator. So they instructed you to take the rate to be a function of theta.

Yeah, but since theta doesn't appear, that means the answer will be like b) for every theta, right? Cause theta gets canceled out in the equations?

iamnotsmart said:
Yeah, but since theta doesn't appear, that means the answer will be like b) for every theta, right? Cause theta gets canceled out in the equations?
Yes.

## 1. How is centripetal acceleration related to the Earth's rotation?

Centripetal acceleration is the acceleration that is directed towards the center of a circular motion. In the case of the Earth's rotation, this acceleration is caused by the gravitational force between the Earth and objects on its surface.

## 2. What is the value of centripetal acceleration on Earth?

The value of centripetal acceleration on Earth depends on the radius of the Earth's rotation and the speed at which it rotates. On average, the Earth's centripetal acceleration is around 0.034 m/s2.

## 3. How does centripetal acceleration affect objects on Earth's surface?

Centripetal acceleration pulls objects towards the center of the Earth, causing them to experience a slight weight increase. However, this effect is so small that it is not noticeable in everyday life.

## 4. Can the Earth's rotation speed change?

Yes, the Earth's rotation speed can change due to various factors such as the gravitational influence of other celestial bodies and changes in the distribution of mass on Earth's surface. However, these changes are very gradual and not noticeable to humans.

## 5. What would happen if the Earth's rotation speed was the same as its escape velocity?

If the Earth's rotation speed was the same as its escape velocity (11.2 km/s), objects on its surface would experience weightlessness as the centripetal acceleration would be equal to the gravitational force pulling them towards the center of the Earth. This scenario is not possible as the Earth's rotation speed is much lower than its escape velocity.

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