How Fast Do Two Identical Spheres Travel When They Meet in Space?

  • Thread starter Thread starter Psyguy22
  • Start date Start date
  • Tags Tags
    Spheres
AI Thread Summary
Two identical 28.5-kg spheres, initially at rest and 35.3 cm apart in space, are analyzed for their speed upon contact due to gravitational attraction. The gravitational potential energy and kinetic energy equations are applied to find the speed when they meet. The initial calculations yielded an incorrect speed of 6.131 x 10^-5 m/s, prompting a review of the distance values used for calculations. The correct distance for r1 is simply the initial separation of 0.353 m, while r2 is 0.242 m at the moment of contact. The weak gravitational force explains the relatively low velocity at impact.
Psyguy22
Messages
62
Reaction score
0

Homework Statement


Two identical 28.5-kg spheres of radius 12.1 cm are 35.3 cm apart (center-to-center distance) and at rest in outer space. (You can assume that the only force acting on each mass is the gravitational force due to the other mass.)

a) If they are released from rest and allowed to fall toward one another, what is their speed when they first make contact?


Homework Equations


ΔE=ΔK+ΔU=0
K=.5mv^2
U=-Gm1m2/r

The Attempt at a Solution


Well ΔK would equal
M*v_i ^2-M*v_f ^2 (because both are moving the 1/2's cancel out)
and ΔU is
-Gm_1 m_2 / r_i - (-Gm_1 m_2 / r_f)

So solved for v I get
v= sqrt(-G m_2 / r_1 + G m_2 / r_2)
which came out to be 6.131*10^-5 m/s which isn't right... I don't see where I went wrong/
 
Physics news on Phys.org
Psyguy22 said:

Homework Statement


Two identical 28.5-kg spheres of radius 12.1 cm are 35.3 cm apart (center-to-center distance) and at rest in outer space. (You can assume that the only force acting on each mass is the gravitational force due to the other mass.)

a) If they are released from rest and allowed to fall toward one another, what is their speed when they first make contact?


Homework Equations


ΔE=ΔK+ΔU=0
K=.5mv^2
U=-Gm1m2/r

The Attempt at a Solution


Well ΔK would equal
M*v_i ^2-M*v_f ^2 (because both are moving the 1/2's cancel out)
and ΔU is
-Gm_1 m_2 / r_i - (-Gm_1 m_2 / r_f)

So solved for v I get
v= sqrt(-G m_2 / r_1 + G m_2 / r_2)
which came out to be 6.131*10^-5 m/s which isn't right... I don't see where I went wrong/

I can not see either, if you do not show your work in detail. What have you used for r1 and r2?

ehild
 
For r1 I have .121+.353+.121 which is .595 m
for r2, it's just .242 m
 
your value for r1 should just be .353 (like point masses) and you're correct that r2 is .242 (distance between centers of mass at collision). The spheres will not have much velocity, as gravity is a very weak force, so it shouldn't be surprising that the value is relatively small
 
Psyguy22 said:
For r1 I have .121+.353+.121 which is .595 m
for r2, it's just .242 m

The center-to center distance was given as 0.353 m , so r1=0.353.


ehild
 
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 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
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...

Similar threads

Back
Top