Hi! I think I see where the r12^3 came from:
r12 ⇒unit vector
⇒r12/(r12)
F21 = -[Gm1m2/(r21)^2]r21
⇒-[Gm1m2/(r21)^2] ⇒(-Gm1m2)(r21)/(r21)^3
since r21 =-r12 and r21 = r12
therefore,
F21 = -Gm1m2/(r12)^2 [-r12/(r12)]
⇒Gm1m2 (r12)/ (r12)^3
⇒Gm1m2/(r12)^2
and since Newton's 3rd law states: F12 =...
Now, let ##\vec{F}_{21}## represent the force that body 2 exerts on body 1, and let ##\vec{r}_{21}## represent the position vector drawn from body 2 to body 1. Then:
$$\vec{F}_{21}=-\frac{Gm_1m_2}{r^3_{21}}\vec{r}_{21}$$
where ##r_{21}## is the magnitude of the position vector ##\vec{r}_{21}##...
Hi, since the gravitational force on me is toward the center of the earth, and is pointing in the downward direction and opposite to the direction of the radius vector, if I designate the downward direction as positive, then how is the direction of the force vector negative? Do we always...
Homework Statement
I am currently reading about Newton's Law of Universal Gravitation and I am so confused as to why there is a negative sign in front of the equation Fg = (G* m1m2)/r^2.
Homework Equations
Fg = (G* m1m2)/r^2
There is a vector form of the magnitude of the gravitational force...
So, is it correct to say the "R" in the left side of equation (Gmems)/R^2 is the radius and the distance between the centers of the satellite and the earth?
Hi, I may be wrong but I thought the "R" on the left side is the distance between the centers of the masses of the satellite and the earth, so the "4R" would only matter in the right side of the equation. Wait, is the distance between the centers of the masses of the satellite and the Earth =...
Homework Statement
A 1,000 kg satellite traveling at speed "v" maintains an orbit of radius, "R" around the earth. What should be its speed if it is to develop a new orbit of radius "4R" ?
Homework Equations
Gravitational force equation (F) = (Gmems)/r^2
G (universal gravitational constant) =...
Since the wheels are sliding, it is assumed to have kinetic friction and so I thought it would be any constant value, not a maximum. I thought maximum frictional force referred to static friction, and since the wheels are sliding for this problem, I thought only kinetic friction mattered.
Hi, I'm not sure but I'm guessing since angular acceleration is Δω/Δt and angular speed is Δθ/Δt, if θ is increasing, angular speed is increasing; therefore, angular acceleration must also be increasing. So, I'm guessing at θ = 60°, the wheels will have the greater angular acceleration. Sorry, I...
Hi, for this particular problem, the assumption is that the wagon is treated as a block "sliding" down the ramp; therefore, like you have mentioned fk (kinetic friction) = μ * n can be used. I thought I can pick a random value for acceleration and plug it into the equation to see the...
Homework Statement
A 20 kg wagon is released from rest from the top of a 15 m long plane, which is angled at 30° with the horizontal. Assuming there is friction between the ramp and the wagon, how is this frictional force affected if the angle of the incline is increased?
Homework Equations...
Hi, so the "distance covered" is 2*3.14*(0.2 m) = 1.3 m and finding "t" through v = 2 π r / t , yielding the answer t = 0.16 s and knowing that the ball makes one revolution in
0.16 s gives the answer (1 revolution / 0.16s ) = 6.3 revolutions / second. Initially I thought I had to divide 1.3 m...
Hi, so according to this equation 1/T, the ball makes an "x" number of revolutions per second. Then, from above I found that T = 0.16 s, therefore
1/0.16 s = 6.3 revolutions / second. How did you get the equation 1/T for the number of revolutions/second? Thanks for your help!
Homework Statement
A 1 kg ball with a radius of 20 cm rolls down a 5 m high inclined plane. Its speed at the bottom is 8 m/s. How many revolutions per second is the ball making when at the bottom of the plane?
Homework Equations
circumference = 2πr
velocity = distance / time = circumference /...