Conservation of energy problem

AI Thread Summary
To determine the velocity a projectile must have at the center of the Earth to reach the surface, one must consider the gravitational potential inside the Earth, which differs from the potential outside. The initial calculation of 11.8 km/s is incorrect; the correct velocity is 7.9 km/s due to the gravitational force being zero at any point inside the Earth. The gravitational potential energy formula must account for the mass distribution within the Earth, leading to a different potential energy calculation. The negative sign in the gravitational potential energy equation indicates that energy is being released as the projectile moves towards the surface, and it is cancelled out by the conservation of energy principle. Understanding these concepts is crucial for accurately solving the problem.
Airsteve0
Messages
80
Reaction score
0
I am having trouble with the following problem:

With what velocity should a projectile be started ar the center of the Earth to reach the surface of the Earth?

When I attempt to solve this using 1/2mv^2 = -(GMm)/R I get an answer of 11.8 km/s. However, the correct answer is 7.9km/s. I know that it has something to do with the potential inside the Earth but I don't even know where to start with incorporating that. Also could someone tell me how the negative on the right hand side is cancelled. Thanks a lot!
 
Physics news on Phys.org
Airsteve0 said:
With what velocity should a projectile be started ar the center of the Earth to reach the surface of the Earth?
At any point r < R inside the earth, the net gravitiational force from all the mass from the radius at that point to the surfce (from r to R) cancels out and is zero. I'm assuming you're supposed to figure out the gravitational potential formula for a point r within the Earth (assuming it's a sphere).

Also could someone tell me how the negative on the right hand side is cancelled.
It's supposed to be negative, and absent any other forces, given an initial condition (position and velocity), the sum of gravitational potential energy and kinetic energy will be a constant. Note that GPE becomes less negative as R increases.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top