Question

Calculate the speed of a satellite in a circular orbit near the Earth (just above the atmosphere). If the mass of the satellite is $\mathbf{200}\mathbf{}\mathbf{kg}$, what is the minimum energy required to move the satellite from this near-Earth orbit to very far away from the Earth?

Short answer

The speed of a satellite is $4.58\times {10}^{-8}\mathrm{m}/\mathrm{s}$

The minimum energy required to move the satellite is $-6.26\times {10}^{-11}\mathrm{J}$

Step-by-step solution

Identification of given data

The mass of the satellite is $200\mathrm{kg}$

Concept of energy in the satellite

A satellite revolves around the earth and it requires energy. This is called as orbiting energy. As the satellite does revolve around the earth, the satellite get kinetic energy in a gravitational field and it gets potential energy.

Calculation of the speed of a satellite

The speed of a satellite,

$v=\sqrt{\frac{GM}{R}}\cdots \cdots \left(1\right)$

Where,

$G=6.6743\times {10}^{-11}\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2$

$M=200\mathrm{kg}$

$R=\text{Radius of the earth}=6371000\mathrm{m}$

Substitute these values in Equation (1),

$v=\sqrt{\frac{GM}{R}}$

$=\sqrt{\frac{6.6743\times {10}^{-11}\times 200}{6371000}·\left(\frac{\left(1\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2\right)\times 1\mathrm{kg}}{1\mathrm{m}}\right)}$

$=\sqrt{\frac{6.6743\times {10}^{-11}\times 200}{6371000}·\left(\frac{(1\mathrm{N}·\mathrm{m})}{\mathrm{kg}}\times \frac{1\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}}\right)}$

$=\sqrt{\frac{6.6743\times {10}^{-11}\times 200}{6371000}·\left(1{\mathrm{m}}^2/{\mathrm{s}}^2\right)}$

$=4.58\times {10}^{-8}\mathrm{m}/\mathrm{s}$

Hence, the speed of a satellite in a circular orbit near the Earth is $4.58\times {10}^{-8}\mathrm{m}/\mathrm{s}$

Calculation of the minimum energy required to move the satellite

Total Energy = Kinetic Energy + Potential Energy

$\text{TE KE + PE}$

$\frac{GMm}{2r}-\frac{GMm}{r}$

$=-\frac{GMm}{2r}$

Substitute these values in above expression,

$TE=-\frac{6.6743\times {10}^{-11}\left(\mathrm{N}·{\mathrm{m}}^2/{\mathrm{kg}}^2\right)\times 5.972(\mathrm{kg})\times {10}^4\times 200(\mathrm{kg})}{2\times 6371000\mathrm{m}}$

$=-6.26\times {10}^{-11}·\left(\frac{\left(\frac{1\mathrm{N}·{\mathrm{m}}^2}{\mathrm{kg}}\right)\times (1\mathrm{kg})\times (1\mathrm{kg})}{1\mathrm{m}}\right)$

$=-6.26\times {10}^{-11}·\left(1\mathrm{N}·\mathrm{m}\times \frac{1\mathrm{J}}{1\mathrm{N}·\mathrm{m}}\right)$

$=-6.26\times {10}^{-11}\mathrm{J}$

The total energy of the satellite is negative which means a satellite will never be able to escape from the gravitational pull of the Earth.