Gravitational Force on Object Stationary to Earth's Surface

[semc]

Show that an object which is stationary with respect to the surface of the Earth is experiencing a gravitational force Ft=(-gmR/r)-[m(w*(w*R))] where R is a vector pointing from the centre of Earth to a position on the surface of the Earth, r is the radius of Earth and w is the angular velocity.

i recognise that [m(w*(w*R))] is the coriolis force but i have no idea which direction it is pointing at. Is it suppose to point to the centre of the Earth?

By the way can someone explain what is angular velocity? I know that its the rate of change of radian per unit time but in 3D they said you can use the right hand rule to find the direction of the angular velocity? What does that mean? Isn't angular velocity just the change of radian per unit time? Why is there a direction?I can't seem to link it up with the fact that it is a vector quantity.

 

[D H]

Show that an object which is stationary with respect to the surface of the Earth is experiencing a gravitational force Ft=(-gmR/r)-[m(w*(w*R))] where R is a vector pointing from the centre of Earth to a position on the surface of the Earth, r is the radius of Earth and w is the angular velocity.

The part of that expression due to gravity is not correct. What does your assignment really say?

i recognise that [m(w*(w*R))] is the coriolis force but i have no idea which direction it is pointing at. Is it suppose to point to the centre of the Earth?

Hint: What if the Earth was rotating much faster than it is?

By the way can someone explain what is angular velocity? I know that its the rate of change of radian per unit time but in 3D they said you can use the right hand rule to find the direction of the angular velocity? What does that mean? Isn't angular velocity just the change of radian per unit time? Why is there a direction?I can't seem to link it up with the fact that it is a vector quantity.

Imagine you are in a car driving along at 60 MPH in the northbound lane of a freeway. 60 MPH is your speed. If your direction is anything but north you are in a heap of trouble. There's a big difference between 60 MPH heading north and 60 MPH heading south. Velocity has magnitude and direction.

The same goes for angular velocity in 3D. In 2D, all you have to worry about is clockwise versus counterclockwise. In 3D, things get a bit hairier. A rotating 3D object is rotating at some angular rate (a magnitude) about some directed axis. First the axis aspect of rotation, sans the direction. Just as there is obviously a difference between a north-south freeway versus an east-west one, rotation about the X axis and is not the same as rotation about the Y axis.

Now the directed aspect. Suppose you are driving along on a north-south freeway. The freeway constrains your motion to a line, but you are still free to go north or south. There is a sign ambiguity. Knowing the axis of rotation brings the problem back to the 2D rotation problem. There is a sign ambiguity problem in 2D, and we solve that with the notion of clockwise versus counterclockwise. The solution to the sign ambiguity problem for 3D rotation is the right hand rotation rule.

Suppose you know the axis about which an object is rotating and the sense in which it rotating. Imagine that the axis as a thin rigid rod that extends out of the object. Curl your the fingers, but not the thumb, of your right hand about this rod so that your fingers are curling in the same sense the object is rotating. Orient your thumb so it is points along the rod. The direction in which your thumb is pointing resolves the sign ambiguity problem mentioned above.

 

[semc]

Wow thanks man i guess i have a better idea about what angular velocity is. So basically your thumb just gives you the axis of rotation if you are going in a circle?

The whole text is like cut and paste from my question what is the problem??

Earth rotates faster means we will get toss into space? I tried to find the direction of the Coriolis force using the right hand rule and if a man is standing on the equator he will experience a gravitational force to the Earth and a Coriolis force orthogonal to the gravitational force?

 

[D H]

The whole text is like cut and paste from my question what is the problem??

The problem is that -GmMR/r term doesn't even make sense as a force. It has units of mass*length³/time². Force has units of mass*length/time². Read the problem again. You are missing a power somewhere.

Earth rotates faster means we will get toss into space? I tried to find the direction of the Coriolis force using the right hand rule and if a man is standing on the equator he will experience a gravitational force to the Earth and a Coriolis force orthogonal to the gravitational force?[/QUOTE]
No, it is not orthogonal.

Have you been taught about the vector cross product yet? What grade level of school are you in? That information will help us homework helpers answer the question in terms of what you are expected to understand.

BTW, this problem is not about the Coriolis force. It is about centrifugal force. Big difference. The Coriolis force on someone standing still on the surface of the Earth is zero. The Coriolis force arises because of velocity. The centrifugal force depends only on position.

 

[semc]

Since R is the vector and r is the magnitude R/r gives us the unit vector? Why ain't the force orthogonal? Yap cross product but i don't really know how to use it for the Coriolis force. I am in uni take physics major. Ain't [m(w*(w*R))] the Coriolis force?

 

[D H]

Since R is the vector and r is the magnitude R/r gives us the unit vector?

Yes, that gives you a unit vector. Does G*M*m have units of force?

Ain't [m(w*(w*R))] the Coriolis force?

No. Google the terms centrifugal force and Coriolis force.

 

[semc]

So essentially its just mv^2/r? Sorry i don't understand the units that you are talking about. I did some dimension analysis and everything seems alright.

 

[D H]

What is Newton's law of gravity?

 

[willem2]

The problem is that -GmMR/r term doesn't even make sense as a force. It has units of mass*length³/time². Force has units of mass*length/time². Read the problem again. You are missing a power somewhere.

I think that g, the acceleration of gravity on the surface of the Earth is meant,
and not G, the gravitational constant. -gm(R/r) is then correct force of gravity.

 

[berkeman]

I think that g, the acceleration of gravity on the surface of the Earth is meant,
and not G, the gravitational constant. -gm(R/r) is then correct force of gravity.

If g is 9.8m/s^2, then why do you need the 1/r term?

 

[willem2]

If g is 9.8m/s^2, then why do you need the 1/r term?

to make R/r a unit vector pointing away from the center of the earth.