Momentum and acceleration due to gravity

AI Thread Summary
The discussion centers on the relationship between momentum, gravity, and danger. Objects with greater momentum pose a higher risk due to their velocity, making them more dangerous than stationary objects. The correct formula for gravitational force is Fg = G m1m2/r^2, and when the radius is doubled, the acceleration due to gravity (g) becomes one-fourth of its original value. It is clarified that Fg is not the same as g; rather, g is derived from the gravitational force equation. Overall, understanding these principles is crucial for assessing danger in various scenarios.
pharaoh
Messages
47
Reaction score
0
what object is dangerous for the target, the one that has greatest momentum or the one that has least momentum and why? what is safest?
I answeres all the parts that contain momentum and velocity and force but i stucked in this one

also
what is the value of g (acceleration due to gravity) if the radius doubled?
is the right formula for this question is Fg= G m1m2/r^2, or do i have to substitute the Fg with m*a, then do the rest
 
Physics news on Phys.org
pharaoh said:
what is the value of g (acceleration due to gravity) if the radius doubled?
is the right formula for this question is Fg= G m1m2/r^2, or do i have to substitute the Fg with m*a, then do the rest

Things with momentum are dangerous because it means they're moving. Are you more scared of a knife on a table or a knife thrown at you?
That equation you posted is correct. Everything should factor out leaving 1/(r^2). How does the gravity change when the r in 1/(r^2) is changed from 1 to 2, or from 2 to 4?
 
so you mean that Fg is the same as g right
and the object that has greatest momentum is the most dangerous
 
if i doubled the radius then the g= 1/4
 
pharaoh said:
so you mean that Fg is the same as g right

No, Fg is not the same as g. From the equation, F=(Gm1m2)/r^2 which gives m2g=(Gm1m2)/r^2 and so g=Gm1/r^2

and the object that has greatest momentum is the most dangerous

Correct

pharaoh said:
if i doubled the radius then the g= 1/4

Right idea: the new value of g obtained when doubling the radius will be 1/4 times the old value of g. So, write g_{new}=\frac{1}{4}g
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Back
Top