Does Mass Affect the Orbital Velocity of Satellites?

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SUMMARY

The discussion focuses on the relationship between mass and orbital velocity for satellites in circular orbits around Earth. It establishes that the orbital velocity, represented by the equation v = √(Gm/r), is independent of the satellite's mass. Specifically, when comparing two satellites, one with mass m and another with mass 3m, both require the same orbital velocity v, as the gravitational force and centripetal force balance each other out regardless of mass. This conclusion is derived from Newton's law of gravitation and the principles of circular motion.

PREREQUISITES
  • Understanding of Newton's law of gravitation
  • Familiarity with circular motion and centripetal force
  • Basic knowledge of gravitational constant (G)
  • Ability to manipulate algebraic equations
NEXT STEPS
  • Study the derivation of the gravitational force equation F = G(m1*m2/r^2)
  • Learn about the implications of mass on gravitational interactions
  • Explore the concept of orbital mechanics and Kepler's laws
  • Investigate the effects of altitude on satellite velocity and orbital period
USEFUL FOR

Students in physics, aerospace engineers, and anyone interested in understanding satellite dynamics and orbital mechanics.

scientict
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Homework Statement



A science group put in a satellite of mass m kg into a circular Earth orbit of radius r. The orbital velocity it needs to remain in this orbit is v. They now put another satellite into a similar orbit at the same altitude. Its mass is 3 times m. What orbital velocity would it need to be given? Give reasons using mathematical reasoning.

Homework Equations



Newton's law v = 2pi x r/T

The Attempt at a Solution



v^2/r=G m/r^2
Satellite 1:
v=√(G m/r)
Satellite 2:
v=√(G 3m/r)
 
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scientict said:

The Attempt at a Solution



v^2/r=G m/r^2

What does 'm' represent here? Can you derive this formula?

When a body moves with constant v in a circular orbit, the centripetal force is constant and equal to the force due to Earth's gravity on the body. Now you can deduce the correct version of the above formula using this.
 

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