Satellite Collision: Find Max Mass mj for Orbit Without Hitting Earth

AI Thread Summary
The discussion revolves around calculating the maximum mass of space junk (mj) that a satellite can collide with while remaining in orbit without hitting Earth. The problem utilizes conservation of angular momentum and energy principles to derive equations for the satellite's velocity before and after the collision. The resulting expression for mj is derived as mj = (√3 - 1)*m, where m is the mass of the satellite. The user seeks confirmation on the correctness of their approach before final submission. The calculations and methodology appear sound based on the provided equations and principles.
jhoge
Messages
15
Reaction score
0

Homework Statement



A satellite of mass m is in a circular orbit of radius 2RE around the Earth. The satellite hits a piece of space junk of mass mj. The space junk, which was initially at rest, sticks to the satellite. The satellite and junk then fall as one in a new elliptic orbit around the Earth. What is the maximum value of the junk mass mj such the satellite of m is to remain in orbit without hitting the Earth? Write your expression in terms of m and no other variables or constants.

Neglect the Earth's atmosphere. Explicitly indicate multiplication using the * symbol

Homework Equations



The equations that are relevant to this problem are the conservation of angular momentum and the conservation of energy equations.

In this case the conservation of angular momentum for the satellite before the collision is :
m*Vo*2R. (Vo is the initial orbital velocity of the satellite. Considering the centripetal
force acting on the satellite we can determine its orbital velocity to be √(G*M/R) At aphelion, the angular momentum of the combined mass of satellite and space junk is:
(m+mj)Va*2R (I am considering the collision to occur at aphelion as this is the farthest point
the satellite can reach from Earth. The satellite slows upon collision, resulting in the elliptical orbit.

At perihelion, the angular momentum is:
(m+mj)Vp*R

The Attempt at a Solution



My approach involves using the conservation of energy and momentum equations mentioned above. In short, I used to conservation of angular momentum to say that

(1) m*Vo*2R = (m+mj)Va*2R &
(2) m*Vo*2R = (m+mj)Vp*R

Using the conservation of energy equation (energy is only conserved after the collision since it is inelastic):

(3) (1/2)(m+mj)(Vp)^2 - GM(m+mj)/R = (1/2)(m+mj)(Va)^2 - GM(m+mj)/2R

I get : (3) Vp^2 - Va^2 = GM/R

Solving for Vp using equation (1) i get Vp = 2*m*√(G*M/R)/(m+mj)
Solving for Va using equation (2) i get Va = m*√(G*M/R)/(m+mj)

Plugging these values into (3) i get

(4) 4m^2Vo^2/(m+mj)^2 - m^2Vo^2/(m+mj)^2 = G*M/R.

Given that Vo = √(G*M/R), equation (4) simplifies to 3m^2 = (m + mj)^2, which leaves me with a value of

mj= (√(3)-1)*m.

I would like to know if this approach is the correct one, because I have one submission attempt left. Thanks in advance.
 
Last edited:
Physics news on Phys.org
forgot to include photo

here is a photo
 

Attachments

  • satellite collision.jpg
    satellite collision.jpg
    11.9 KB · Views: 466
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 .
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...
Back
Top