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(a) Determine the amount of work (in joules) that must be done on a 95 kg payload to elevate it...

Question:

(a) Determine the amount of work (in joules) that must be done on a 95 kg payload to elevate it to a height of 1007 km above the Earth's surface. ANSWER in MJ

(b) Determine the amount of additional work that is required to put the payload into circular orbit at this elevation. ANSWER in J

The energy of a Satellite:

{eq}\\ {/eq}

Any object that orbits around another celestial body under the influence of the gravitational pull of that celestial body, is called a satellite of that celestial body. Once a satellite is launched into an orbit, the motion of the satellite preserves naturally, that is, there is no driving force required to keep the motion of the satellite intact.

The mechanical energy of a satellite, therefore, remains constant throughout its orbit. The energy of a satellite consists of its kinetic energy and the gravitational potential energy between the satellite and the planet.

Both of them depends only on the following factors:

  1. The mass of the satellite.
  2. The mass of the planet.
  3. The distance of the satellite from the center of the planet.

Answer and Explanation:

{eq}\\ {/eq}

We are given:

  • The mass of the payload, {eq}M=95\;\rm kg {/eq}
  • The change in the height of the payload above the Earth's surface, {eq}\Delta h=1007\;\rm km=1.007\times 10^{6}\;\rm m {/eq}

a)

The amount of work required to raise an object from the surface of Earth is equal to the change in the gravitational potential energy of the object.

The gravitational potential energy between two-point sized masses {eq}m_1 {/eq}, and {eq}m_2 {/eq}, depends on the separation between them by the equation:

{eq}U=-\dfrac{Gm_1m_2}{r} {/eq}

Here,

  • {eq}G=6.67\times 10^{-11}\;\rm J.m/kg^2 {/eq} is the gravitational constant.


The mass of Earth is: {eq}M=5.98\times 10^{24}\;\rm kg {/eq}

The radius of Earth is: {eq}R=6.4\times 10^{6}\;\rm m {/eq}

A spherically shaped mass ( with a uniform density) behaves like a point mass of the same magnitude as the original object, located at the center of the sphere.

Therefore, the initial distance between the payload and the Earth is equal to the radius of the Earth.

{eq}r_1=R=6.4\times 10^{6}\;\rm m {/eq}

The final distance between the payload and the Earth is:

{eq}r_2=R+\Delta h=6.4\times 10^{6}\;\rm m+1.007\times 10^{6}\;\rm m=7.407\times 10^{6}\;\rm m {/eq}


The change in the gravitational potential energy is:

{eq}\begin{align*} \Delta U&=U_f-U_i\\ &=\dfrac{-GMm}{r_2}-\dfrac{-GMm}{r_1}\\ &=GMm\left ( \dfrac{1}{r_1}-\dfrac{1}{r_2} \right )\\ &=6.67\times 10^{-11}\;\rm J.m/kg^2 \times 5.98\times 10^{24}\;\rm kg\times 95\;\rm kg \left ( \dfrac{1}{6.4\times 10^{6}\;\rm m}-\dfrac{1}{7.407\times 10^{6}\;\rm m} \right )\\ &=3.789\times 10^{16}\;\rm J.m \times \left ( 1.5625\times 10^{-7}\;\rm m^{-1}-1.35\times 10^{-6}\;\rm m^{-1} \right )\\ &=3.789\times 10^{16}\;\rm J.m\times 2.125\times 10^{-7}\;\rm m^{-1}\\ &=8.05\times 10^{9}\;\rm J\\ &=8.05\times 10^{3}\times 10^{6}\;\rm J\\ &=\boxed{8.05\times 10^{3}\;\rm MJ} \end{align*} {/eq}


b)

The total energy of a satellite depends only on the mass of the satellite, {eq}m {/eq}, the mass of the planet/star, {eq}M {/eq}, orbited by the satellite, and the radius, {eq}r {/eq}, of the circular orbit followed by the satellite by the following equation:

{eq}E=\dfrac{-GMm}{2r} {/eq}

After plugging the given values into the above equation, we have:

{eq}\begin{align*} E&=\dfrac{-6.67\times 10^{-11}\;\rm J.m/kg^2 \times 5.98\times 10^{24}\;\rm kg \times 95\;\rm kg}{2\times 7.407\times 10^{6}\;\rm m}\\ &=\dfrac{-3.789\times 10^{16}\;\rm J.m }{14.814\times 10^{6}}\\ &=-2.56\times 10^{9}\;\rm J \end{align*} {/eq}


The gravitational potential energy at the final location is:

{eq}\begin{align*} U_f&=-\dfrac{GMm}{r_2}\\ &=-\dfrac{-6.67\times 10^{-11}\;\rm J.m/kg^2 \times 5.98\times 10^{24}\;\rm kg \times 95\;\rm kg}{7.407\times 10^{6}\;\rm m}\\ &=-5.12\times 10^{9}\;\rm J \end{align*} {/eq}


The additional energy required is:

{eq}\begin{align*} \Delta E&=E-U_f\\ &=-2.56\times 10^{9}\;\rm J-\left (-5.12\times 10^{9}\;\rm J \right )\\ &=\boxed{2.56\times 10^{9}\;\rm J} \end{align*} {/eq}



Learn more about this topic:

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Stable Orbital Motion of a Satellite: Physics Lab

from Physics: High School

Chapter 8 / Lesson 16
565

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