A (part 1 of 3) 10.0 points Given: Mearth = 5.98 × 1024 kg Rearth = 6.37 × 106 m
ID: 1397517 • Letter: A
Question
A (part 1 of 3) 10.0 points
Given:
Mearth = 5.98 × 1024 kg Rearth = 6.37 × 106 m
A satellite of mass 988 kg is in a circular or- bit at an altitude of 523 km above the earth’s surface. Because of air friction, the satellite eventually is brought to the earth’s surface, it hits the earth with a velocity of 5 km/s. Let the gravitational potential energy be zero at r = . The universal gravitational constant G = 6.67259 × 1011 N m2/kg2.
What is the total energy of the satellite in orbit?
Answer in units of J.
B (part 2 of 3) 10.0 points
What is the total energy of the satellite just before it hits the ground?
Answer in units of J.
C (part 3 of 3) 10.0 points
What is the work done by friction? Answer in units of J.
Explanation / Answer
Given that,
Me = 5.98 x 1024 Kg ; Re = 6.37 x 106 m
m = 998 Kg ; A = 523 km = 523 x 103 m; v = 5 km/s = 5000 m/s
A)We need to find the total energy of satellite in orbit. The total energy would be the sum of its KE and PE.
E(orbit) = -G Me m / Re + A + 1/2 m v2
As the centripital force required by satellite is given by the gravitational force, so
-G Me m / (Re + A)2 = m v2 / (Re + A)
We eliminate v from the E(orbit) using the above relation and get:
E(orbit) = -1/2 [G Me m / (Re + A)]
E(orbit) = - 0.5 x 6.67259 × 10-11 x 5.98 x 1024 x 988 / (6.37 x 106 + 523 x 103 )
E(orbit) = - 19711.631571 x 1013 / 6893000 = 28596593023.357029
E(orbit) = - 2.86 x 1010 Joules
(b)The total energy is the sum of KE + PE
E(total- ground) = - G Me m / Re + 1/2 m v2
E(total- ground) = - 6.67259 × 10-11 x 5.98 x 1024 x 988 / 6.37 x 106 + 0.5 x 988 x (5000)2
E(total- ground) = -61888953126.531 + 12350000000 = -49538953126.531
hence, E(total- ground) = -4.9 x 1010 Joules
(c)We need to find the work done by the friction.
From work energy theorem, its the diffrence of intial and final energies. So
W(friction) = E(total- ground) - E(orbit)
W(friction) = -4.9 x 1010 - (- 2.86 x 1010 ) = -2.04 x 1010 Joules
Hence, W(friction) = -2.04 x 1010 Joules
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.