Calculating Relative Acceleration and Velocity with Newton's Law of Gravity

Click For Summary
SUMMARY

This discussion focuses on calculating the relative acceleration and velocity of two particles with masses m and M, which attract each other according to Newton's law of gravity. The derived formula for their relative velocity of approach is v(rel) = √(2G(M+m)/a). The particles start at rest from an infinite distance, and their center of mass remains stationary throughout the motion. The conservation of energy principle is applied, leading to the conclusion that the mutual potential energy at distance 'a' is U(pot) = -G(mM/a^2).

PREREQUISITES
  • Understanding of Newton's law of gravity
  • Basic principles of conservation of momentum and energy
  • Familiarity with potential energy concepts in gravitational systems
  • Knowledge of algebraic manipulation to solve equations
NEXT STEPS
  • Study the derivation of gravitational potential energy in detail
  • Learn about the conservation of momentum in isolated systems
  • Explore the implications of relative motion in gravitational fields
  • Investigate advanced applications of Newton's laws in multi-body systems
USEFUL FOR

Students of physics, astrophysicists, and anyone interested in gravitational interactions and dynamics of particle systems.

starbaj12
Messages
48
Reaction score
0
Two particles having masses of m and M respectively, attract each other according to Newton's law of gravity. Initially they are at rest at an infinite distance apart. Find their relative acceleration and show that their relative velocity of approach is

v(rel) = square root(2G(M+m)/a)

I do not even know where to start on this problem, if any could give me help that would be great.

Thank you
 
Physics news on Phys.org
Momentum and energy are conserved!
 
starbaj12 said:
Two particles having masses of m and M respectively, attract each other according to Newton's law of gravity. Initially they are at rest at an infinite distance apart. Find their relative acceleration and show that their relative velocity of approach is

v(rel) = square root(2G(M+m)/a)

I do not even know where to start on this problem, if any could give me help that would be great.

Thank you

The particles are initially at rest so their angular momentum is zero, they will move towards each other along the straight line connecting them.
As there are no external forces acting on this system of particles, their centre of mass stays steady during the motion. You can place the origin of the frame of reference there. That is

[tex]mv+MV=0\mbox{ } \rightarrow \mbox{ }V=-\frac{M}{m}v[/tex]

The energy of the system is conserved, and it is zero, as the particles started from rest at infinite distance apart.
At distance 'a' the mutual potential energy is

[tex]U_{pot}= - G\frac{mM}{a^2}[/tex].

[tex]\frac{1}{2}mv^2+\frac{1}{2}MV^2-G\frac{mM}{a^2}=0[/tex]

You have two equations for v and V. Solve, and determine their relative speed [tex]v_{rel}=|v-V|[/tex].

ehild
 
  • Like
Likes   Reactions: ShayP

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 24 ·
Replies
24
Views
2K
  • · Replies 51 ·
2
Replies
51
Views
6K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 117 ·
4
Replies
117
Views
10K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 14 ·
Replies
14
Views
3K