A satellite of mass 988 kg is in a circular or- bit at an altitude of 523 km abo

Question

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,

006 (part 1 of 3) 10.0 points Given: Marth 5.98 x 1024 kg Rearth = 6.37 x 106 m earth - 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 - oo. T G - 6.67259 x 10-11 Nm2/kg2 orbit? he universal gravitational constant What is the total energy of the satellite in Answer in units of J. 007 (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. 008 (part 3 of 3) 10.0 points What is the work done by friction? Answer in units of J.

Explanation / Answer

A)

its total energy is the sum of potential energy and kinetic energy

Eorbit = - G*M*m / Re +A + 0.5* m*v0^2

but since the required centripital force is supplied by the gravitational force

G * M * m / (Re + A)^2 = m * v0^2 / Re + A

use the relation to eliminate the v0 in the Eorbit formula then,

Eorbit = - 0.5 * G* M* m / Re +A

then just put the values in the formula

Eorbit = - 0.5 * 6.67259 * 10^-11 * 5.98 * 10^24 * 988 / ( 6.37 * 10^6 + 523000)

Eorbit = - 19711.6316 * 10^13 /6893000 = - 0.002859 * 10^13 J

so the total energy of the satellite in the orbit is -2.86 * 10^10 J

B)

the total energy of the satellite just before it hit the ground is the sum of potential energy and kinetic energy

Eground = - G * M * m / Re + 0.5 * m * v^2

Eground = - 6.67259 * 10^-11 * 5.98 * 10^24 * 988 / 6.37 * 10^6 + 0.5 * 988 * 5000^2

Eground = - 6188.894 * 10^7 + 1.235*10^10

Eground = - 7.424 * 10^10 J

so the total energy before hit the ground is -7.424 * 10^10 J

C)

from the work energy theorem

delta W = Eground - Eorbit

delta W = -7.424 * 10^10 - ( - 2.86 * 10^10 )

delta W = - 4.564 * 10^10 J

so the work done by the friction is 4.564 * 10^10 J

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