Question

For a satellite to be in a circular orbit 890 km above the surface of the earth, (a) what orbital speed must it be given, and (b) what is the period of the orbit (in hours)?

Short answer

a) The orbital speed given to the satellite must be $7.40\mathrm{𝟠}x{10}^{3}m/s$.

b) The period required for completing one revolution/orbit is 1.7 hr.

Step-by-step solution

Identification of given data

The given data can be listed below,

• The height of the satellite above the earth is, $h=890\mathrm{km}\left(\frac{1000\mathrm{mm}}{1\mathrm{km}}\right)$.

Significance of satellite motion

When a satellite circles the earth and it is near the surface, it follows an equipotential surface rule. Gravity has a constant value on an equipotential surface. It implies that while the satellite travels, it moves 'up' and 'down' to the surface, although minor differences exist.

(a) Determination of the orbital speed of the satellite

The radius of the orbit can be given by,

r = h + R

Here, h is the height of the satellite above the earth, and R is the radius of the earth whose value is$6.37x{10}^{6}m$

Substitute the value above equation will give,

$$\begin{gathered} r=637\times {10}^6\mathrm{m}+0.89\times {10}^{6}m \\ =7.26\times {10}^{6}m \end{gathered}$$

From the law of conservation of energy, the expression for force can be given by,

$\frac{G{M}_{E}m}{r^2}=\frac{m{v}^2}{r}$

Here,$M_E$ is the mass of the earth whose value is $5.97\times {10}^{24}\mathrm{kg}$, G is the constant of gravitation whose value is $6.673\times {10}^{-11}N·m^2/{\mathrm{kg}}^2$,m is the mass of the satellite, r is the radius of the orbit, and v is the orbital speed of the satellite.

By solving the above equation,

$v=\sqrt{\frac{G{M}_E}{r}}$

Substitute all values in the above,

$\begin{array}{r}v=\sqrt{\frac{\left(6.673\times {10}^{-11}\mathrm{N}\cdot {\mathrm{m}}^2/{\mathrm{kg}}^2\right)\left(5.97\times {10}^{24}\mathrm{kg}\right)}{7.26\times {10}^6\mathrm{m}}\left(\frac{1\mathrm{kg}\cdot \mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}}\right)}\\ =7.408\times {10}^3\mathrm{m}/\mathrm{s}\end{array}$

Thus, the orbital speed given to the satellite is $7.408\times {10}^3\mathrm{m}/\mathrm{s}$.

(b) Determination of the period of theorbit

The time period of the orbit is given by,

$T=\frac{2\pi r}{v}$

Here, r is the radius of the orbit, and v is the orbital speed of the satellite.

Substitute values in the above equation.

$$\begin{gathered} T=\frac{2\pi \left(7.26\times {10}^6\mathrm{m}\right)}{7.408\times {10}^3\mathrm{m}/\mathrm{s}} \\ =6158s\left(\frac{1hr}{3600s}\right) \\ =1.7hr \end{gathered}$$

Thus, the period required for completing one revolution/orbit is 1.7 hr.