A 1200 kg satellite is orbiting the earth in a circular orbit with an altitude of 1700 km....

Question:

A 1200 kg satellite is orbiting the earth in a circular orbit with an altitude of 1700 km. {eq}\rm (M_{earth} = 5.9742 \times 10^{24} \ kg, \ {/eq} {eq}\rm R_{earth} = 6.3781 \times 10^6 \ m, \ M_{moon} = 7.36 \times 10^{22} \ kg, \ R_{moon} = 1.7374 \times 10^6 \ m, \ D_{earth \ to \ moon} = 3.844 \times 10^8 \ m \ (center \ to \ center), \ {/eq} {eq}\rm \ G = 6.67428 \times 10^{-11} \ N \cdot m^2/kg^2) {/eq}

(a) How much energy does it take just to get it to this altitude?

(b) How much kinetic energy does it have once it has reached this altitude?

(c) What is the ratio of this change in potential energy to the change in kinetic energy?

(d) What would this ratio be if the final altitude of the satellite were 4700 km?

(e) What would this ratio be if the final altitude of the satellite were 3185 km?

Total Mechanical Energy of an Object on the Surface and in an Orbit About Earth

An object on the surface of earth has gravitational potential energy and kinetic energy. The gravitational potential energy of the object of mass m on the surface of earth {eq}U_s = - \dfrac { G M m } { R } {/eq}. Here {eq}G, \ \ M, \ \ R {/eq} are the gravitational constant, mass of the earth and radius of the earth. All objects on the surface of the earth rotates along with the earth and as a result will have some velocity {eq}v = r \omega {/eq}. Here {eq}r, \ \ \omega {/eq} are radial distance from the axis of the earth to the object and angular velocity of the earth about its own axis. This linear velocity provides kinetic energy to all objects on the surface of earth.

When an object of mass m moves in an orbit of radius r, it has gravitational potential energy {eq}U _o = \dfrac { - G M m } { r } {/eq} and kinetic energy {eq}K_o = \dfrac { G M m } { 2 r } {/eq}

Answer and Explanation:

Given data

  • Mass of the satellite {eq}m = 1200 \ kg {/eq}
  • Altitude of the circular orbit of the satellite {eq}h_1 = 1.700 \times 10^6 \ m {/eq}
  • Mass of earth {eq}M = 5.9742 \times 10^{24} \ kg {/eq}
  • Radius of earth {eq}R = 6.3781 \times 10^6 \ m {/eq}
  • Gravitational constant {eq}G = 6.674 \times 10^{-11} \ N . m^2 /kg^2 {/eq}

Part a )

Distance to the orbit from the center of the earth {eq}r_1 = R + h_1 \\ r_1 = 6.3781 \times 10^6 + 1.70 \times 10^6 \\ r_1 = 8.0781 \times 10^6 \ m {/eq}

Potential energy of the satellite in the orbit at the given altitude {eq}U_1 = \dfrac { - G M m } { r_1 } {/eq}

In order to get the satellite to the given altitude we need to supply an energy equal to the difference in potential energy at the orbit and on the surface of earth.

Therefore the energy need to be supplied to the satellite to be taken to the orbit {eq}\Delta E = - G M m \times ( \dfrac { 1 } { r_1 } - \dfrac { 1 } { R } ) \\ \Delta E = - 6.674 \times 10^{-11} \times 5.9742 \times 10^{24} \times 1200 \times ( \dfrac { 1 } { 8.0781 \times 10^6 } - \dfrac { 1 } { 6.3781 \times 10^6 } ) \\ \Delta E = 1.5787\times 10^{10} \ J {/eq}

Part b)

If the satellite has to remain at this altitude in the given orbit, it needs some orbital velocity v that will provide sufficient centripetal force equal to the gravitational force acting on the satellite.

Then the required kinetic energy of the satellite {eq}K_1 = \dfrac { G M m } { 2 r_1 } \\ K_1 = \dfrac { 6.674 \times 10^{-11} \times 5.9742 \times 10^{24 } \times 1200 } { 2 \times 8.0781 \times 10^6 } \\ K_1 = 2.9615 \times 10^{10} \ J {/eq}

Part c)

Ratio of the change in potential energy to change in kinetic energy {eq}\dfrac { \Delta U } { \Delta K } = \dfrac { \Delta E } { K_1 } \\ \implies \dfrac { \Delta U } { \Delta K } = \dfrac { 1.5787 \times 10^{10} } { 2.9615 \times 10^{10} } \\ \implies \dfrac { \Delta U } { \Delta K } = 0.5331 {/eq}

Part d)

Given altitude {eq}h_2 = 4.700 \times 10^6 \ m {/eq}

Radius of the orbit of the satellite {eq}r_2 = R + h_2 \\ r_2 = 6.3781 \times 10^6 + 4.700 \times 10^6 \\ r_2 = 1.10781\times 10^7 \ m {/eq}

Change in gravitational potential energy {eq}\Delta U = - G M m \times ( \dfrac { 1 } { r_2 } - \dfrac { 1 } { R } ) \\ {/eq}

Change in kinetic energy {eq}\Delta K = \dfrac { G M m } { 2 r_2 } {/eq}

Therefore the ratio of change in potential energy to change in kinetic energy {eq}\dfrac { \Delta U } { \Delta K } = - G M m \times \left[ \dfrac { 1 } { r_2 } - \dfrac { 1 } { R } \right] \times \dfrac { 2 r_2 } { G M m } \\ \dfrac { \Delta U } { \Delta K } = - \left[ \dfrac { 1 } { 1.10781 \times 10^7 } - \dfrac {1 } { 6.3781 \times 10^6 } \right] \times 2 \times 1.10781 \times 10^7 \\ \dfrac { \Delta U } { \Delta K } = 1.474 {/eq}

Part e)

Given altitude {eq}h_3 = 3.185 \times 10^6 \ m {/eq}

Radius of the orbit of the satellite {eq}r_3 = R + h_3 \\ r_3 = 6.3781 \times 10^6 + 3.185 \times 10^6 \\ r_3 = 9.5631\times 10^6 \ m {/eq}

Change in gravitational potential energy {eq}\Delta U = - G M m \times ( \dfrac { 1 } { r_3 } - \dfrac { 1 } { R } ) {/eq}

Change in kinetic energy {eq}\Delta K = \dfrac { G M m } { 2 r_3 } {/eq}

Therefore the ratio of change in potential energy to change in kinetic energy {eq}\dfrac { \Delta U } { \Delta K } = - G M m \times \left[ \dfrac { 1 } { r_3 } - \dfrac { 1 } { R } \right] \times \dfrac { 2 r_3 } { G M m } \\ \dfrac { \Delta U } { \Delta K } = - \left[ \dfrac { 1 } { 9.5631 \times 10^6 } - \dfrac {1 } { 6.3781 \times 10^6 } \right] \times 2 \times 9.5631 \times 10^6 \\ \dfrac { \Delta U } { \Delta K } = 0.9987 {/eq}


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Stable Orbital Motion of a Satellite: Physics Lab

from Physics: High School

Chapter 8 / Lesson 16
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