15. How much work must be done to lift an artificial satellite of mass $$ m $$ from the surface of Earth and put it in a circular orbit with a radius equal to twice the Earth's Radius?

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We can solve the problem by computing the change in the satellite's total mechanical energy (gravitational potential energy plus kinetic energy) as it is moved from the Earth's surface to the orbit. Here are the steps:

1. Gravitational Potential Energy (GPE):  
   The gravitational potential energy of a mass m at a distance r from the center of the Earth is given by  
  U = –GMm/r  
   where G is the gravitational constant and M is the mass of the Earth.

2. Kinetic Energy in a Circular Orbit:  
   For a circular orbit of radius r, the satellite’s orbital speed v is given by the condition for circular motion:  
  GMm/r² = m(v²/r)  ⟹ v² = GM/r  
   so the kinetic energy is  
  K = ½ m v² = ½ m (GM/r) = GMm/(2r).

3. Total Mechanical Energy:  
   The total mechanical energy (E) in orbit is the sum of the kinetic and potential energies:  
  E = K + U = (GMm/(2r)) – (GMm/r) = –GMm/(2r).

Now, consider the two positions:

• Initially, the satellite is on the Earth's surface (r = R). It is at rest, so its kinetic energy is zero, and its gravitational potential energy is  
  U_i = –GMm/R.  
   Thus its total energy initially is  
  E_i = –GMm/R.

• Finally, the satellite is in a circular orbit at r = 2R. Its total energy there is  
  E_f = –GMm/(2(2R)) ? 
  Actually, be careful: using the formula derived above, at r = 2R the total energy is  
   E_f = –GMm/(2(2R))  
      = –GMm/(4R).  
   (Alternatively, you can compute the gravitational potential energy U_f = –GMm/(2R) and the kinetic energy K_f = GMm/(4R), so that E_f = K_f + U_f = –GMm/(4R).)

4. Change in Total Mechanical Energy:  
   The work done (W) against gravity to move the satellite into orbit (assuming an ideal process with no losses) is the change in total energy:
  W = E_f – E_i  
    = (–GMm/(4R)) – (–GMm/R)  
    = GMm/R – GMm/(4R)  
    = ( (4GMm – GMm) / (4R) )  
    = 3GMm/(4R).

Thus, the work required is
  W = 3GMm/(4R).

An alternative form uses the fact that the acceleration due to gravity at Earth’s surface is g = GM/R². Then, GM can be written as gR² and
  GM/R = gR.
So the work becomes:
  W = (3/4) gR m.

Both forms are equivalent.

Final Answer: The work required is W = 3GMm/(4R), or equivalently, (3/4) m gR.
The work required to lift the satellite is $$ \frac{3}{4} $$ times the product of the gravitational constant $$ G $$, the Earth's mass $$ M $$, the satellite's mass $$ m $$, and the Earth's radius $$ R $$. Alternatively, it can be expressed as $$ \frac{3}{4} $$ times the satellite's mass $$ m $$, the acceleration due to gravity $$ g $$, and the Earth's radius $$ R $$.

15. How much work must be done to lift an artificial satellite of mass \( m \) from the surface of Earth and put it in a circular orbit with a radius equal to twice the Earth's Radius?

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