Q21E - University Physics with Modern Physics

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)?

Step-by-Step Solution

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 1: Identification of given data

The given data can be listed below,

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

Step 2: 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.

Step 3: (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 $7.408\times {10}^3\mathrm{m}/\mathrm{s}$.

Step 4: (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.

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

 

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