A 55-kg satellite is in circular orbit earth with an orbital radius of 7.4 x 10^6 m. Determine...

Question:

A 55-kg satellite is in circular orbit earth with an orbital radius of 7.4 x 10^6 m. Determine the satellite's:

a) kinetic energy

b) gravitational potential energy

c) total energy

d) binding energy

The Orbital speed of a satellite about Earth:

A satellite that is orbiting the Earth at its orbit has an orbital speed.

The orbital speed of the satellite is given by;

{eq}v = \sqrt{\dfrac{GM}{r}} {/eq}

where,

  • {eq}M {/eq} is the mass of the Earth.
  • {eq}r {/eq} is orbiting radius of the satellite.
  • {eq}G = 6.67 \times 10^{-11} \ Nm^2/kg^2 {/eq} is the gravitational constant.

Answer and Explanation:

Let

  • r is the orbital radius of the satellite.
  • m is the mass of the satellite.
  • M is the mass of the earth.

Given

  • {eq}r =7.4\times 10^6\ m {/eq}
  • Mass of the satellite (m) = 55 kg
  • Mass of the earth (M) ={eq}5.972\times 10^24 \ kg {/eq}

Orbital velocity of the satellite is given by:

{eq}\begin{align} v_0 &= \sqrt{\dfrac{GM}{r}}\\ \end{align} {/eq}

Part (a)

The kinetic energy of satellite is

{eq}\begin{align} K.E. &= \dfrac{1}{2}mv_0^2 \\ &= \dfrac{1}{2}m{\dfrac{GM}{r}}\\ &= \dfrac{1}{2}\times 55\ kg \times \dfrac{6.673\times10^{-11}\ Nm^2kg^{-2}\times 5.972\times 10^{24}\ kg}{7.6\times 10^6}\\ &= 1.44 \times 10^9\ J\\ \end{align} {/eq}

Therefore, the kinetic energy of the satellite is

{eq}1.44 \times 10^9\ J {/eq}.

Part (b)

The gravitational potential energy (P.E.) is

{eq}\begin{align} P.E. &=-\dfrac{GMm}{r}\\ &= -2\times K.E.\\ &= -2.88\times 10^9 \ J\\ \end{align} {/eq}

Part (c)

Total energy of the satellite is;

{eq}\begin{align} E &=K.E. + P.E.\\ &= -1.44 \times 10^9\ J\\ \end{align} {/eq}

Part (d)

Binding energy of the satellite is

{eq}\begin{align} B.E. &= -E\\ \implies B.E. &= 1.44 \times 10^9\ 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
510

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