Question

A 507 kg satellite is in a circular orbit at an altitude of 754 km above a planet’s surface. This planet is similar to our Earth. Because of air friction, the satellite eventually is brought to the Earth’s surface, and it hits the Earth with a speed of 3 km/s. The radius of the planet is 7 × 106 m and its mass is 8 × 1024 kg. The gravitational constant is 6.67259 × 10−11 N m2 /kg2. How much energy was absorbed by the atmosphere through friction? Answer in units of J.

Answer 1

In this question , the initial gravitational potential energy is converted to Kinetic enery and one part of it is lost due to airfriction ,
We have to find the work done by air resistance , which will be thedifference in Initial and final energies
Initial energy = -GMm / r = -6.67*10^-11 *504*8*10²⁴ / (7*10⁶+754*10³)
       U = -2.69*10¹⁷/(7754*10³) =-3.5*10¹⁰ J

Finally Kinetic energy = mv²/2 = 504*3000*3000 /2 =2.27*10⁹ J
   (Assuming the mass to be constant while falling )
   Final Gravitational potential energy =6.67*10^-11 * 504*8*10²⁴ /(7*10⁶)
                                                       = -3.84*10¹⁰ J
Work by air resistance= Difference in initial and finalenergies
                     = initial - Final = (-35 - 2.27 +38.4 )*10⁹ J
   = 1.13*10⁹ J