Artificial Satellites

The Mission: 

Students will be able to explain the concept of thrust of rockets and gravitational binding energy. The students are expected to create a game in which they have to control the thrust of an artificial satellite to avoid collision with asteroids and other satellite 

2. Concepts to Understand

2.1 Thrust 

Thrust is a force that acts as a push to make an object move away from the other object. According to Newton’s third law of motion, every action has equal and opposite reaction. 

 For example, an aircraft engine acts on air and pushes it back. This causes the air to push the aircraft forward. Thus, air reacts on the aircraft to move forward. This reaction force is called as thrust.  

2.2 Thrust in a rocket 

Thrust produced by rocket make them propel forward. The gases formed by burning fuel are ejected at a very high speed. There is a push acting on these gases. Once these gases come out of the rocket, they apply reaction force on the rocket in the direction opposite to that of exhaust. As a result, rocket moves up, away from the earth.  

As no external forces are involved in this, hence the linear momentum of the rocket is said to be conserved. 

Let’s derive a rocket equation. Let’s say at time t, mass of the rocket with the fuel is m, and its initial velocity is v. At time t + 

ΔΔ

t fuel burnt by rocket is 

ΔΔ

m, which is converted into exhaust gases. The relative velocity of these exhaust gases is u. As a result of the exhaust, rocket gains velocity 

ΔΔ

1. Thus, the final velocity becomes v +

ΔΔ

1. v.

Due to loss of fuel, the final mass of rocket at time t + 

ΔΔ

t is m – 

ΔΔ

m 

The velocity of exhaust gases will be v – u. 

According to the law of conservation of linear momentum, we can write 

mv=(m−Δm)(v+Δv)+Δm (v−u)mv=m−Δmv+Δv+Δm v−u

 On solving the above equation, we get 

mv=mv+mΔv−Δmv−ΔmΔv+Δmv−Δmumv=mv+mΔv−Δmv−ΔmΔv+Δmv−Δmu

 Solving the equation and ignoring the term 

ΔmΔvΔmΔv

 as it is negligible, we get, 

mΔv=Δmu           …(1)mΔv=Δmu           …(1)

 Next, dividing both the side by 

ΔΔ

t we get 

m{Δv}{Δt}=u{Δm}{Δt}           …(2)mΔvΔt=uΔmΔt           …(2)

 At time t, mass of the rocket is m, and at time t + 

ΔΔ

t the mass of the rocket is m – 

ΔΔ

m 

Taking the difference in the initial and the final mass we get,  

m(t+Δt)−m(t)= m−Δm−m= −Δmmt+Δt−mt= m−Δm−m= −Δm

 Differentiating the equation 1 and considering the mass difference we get 

mdvdt= −udmdtmdvdt= −udmdt

 T= ma= −udmdt       …(3)T= ma= −udmdt       …(3)

Thus, the thrust produced by the rocket is given by the equation (3). 

On integrating the equation 

mdv= −u dmmdv= −u dm

 for the limits of velocity vi to vf and the limits of mass mi to mf, we get 

m∫vfvidv= −u ∫mfmidmm∫vivfdv= −u ∫mimfdm

On solving the above equation, we get the equation of velocity of rocket as 

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For any rocket to take off the ratio of the thrust produced by it to its total weight should be always greater than 1.  

The thrust to weight ratio is given as: 

TW=mamg=agTW=mamg=ag

The more the acceleration produced by the rocket, the faster it can escape the gravitational pull of the earth. 

2.3 Orbital velocity and gravitational binding energy  

Once rocket lifts any satellite to a certain height, called its parking orbit point, then the satellite needs to start revolving into this orbit with some velocity v. if the satellite fails to do so, then it will crash back to the earth. 

The work done in taking satellite to reach to the height dr, is dW = F x dr. here F is the gravitational force. 

On integrating the equation for the limits of the height dr from h1 to h2 we get 

∫dW= ∫h2h1GMEmr2 dr∫dW= ∫h1h2GMEmr2 dr

W=GMEm×(1h1−1h2)W=GMEm×1h1−1h2

Substituting h1 = R, and h2 = h for h > R, we can substitute for the WR as gravitational potential at the earth i.e. W0 in the above equation we get 

W=GMEmh+W0W=GMEmh+W0

This equation gives the potential energy stored in the rocket at the height h. This energy is then converted into kinetic energy by a small push by the thruster in the satellite itself. 

 T hus the satellite captures the orbit and starts rotating around the earth. 

The centripetal force acting on the satellite is  

Fc=mv2oRe+hFc=mvo2Re+h

And the gravitational force acting on the satellite is  

FG=GMem(Re+h)2FG=GMem(Re+h)2

As Fc = Fg hence we can write 

mv2oRe+h=GMem(Re+h)2mvo2Re+h=GMem(Re+h)2

v2o=GMeRe+hvo2=GMeRe+h

Thus, taking square root on both the side we get 

vo=GMeRe+h−−−−−−√= gRe−−−√vo=GMeRe+h= gRe

This equation gives the orbital velocity of the satellite with which it should move in the orbit at the height h above the Earth’s surface. 

And the binding energy of the satellite becomes 

E=mv2o2+W0=GMemh+W0E=mvo22+W0=GMemh+W0

2.4 Escape velocity 

If a satellite has velocity 

v=2gRe−−−−√= 2GMe/(Re+h) −−−−−−−−−−−−−√v=2gRe= 2GMe/(Re+h) 

 then the satellite does not stays in orbit and it flies away from the earth.