Chapter 7.2, Problem 17PP

a.

To determine

The speed of a satellite that is in orbit above Mercury’s surface.

Answer to Problem 17PP

  $2.85\times {10}^3{\text{m/s}}^2$

Explanation of Solution

Given:

The mass of Mercury is $3.30\times {10}^{23}\text{kg}$ .

The radius of Mercury is $2.44\times {10}^6\text{m}$ .

Height of satellite above mercury’s surface is $260\text{km}=0.26\times {10}^6\text{m}$

Formula used:

Orbital speed of a satellite can be obtained by:

  $v=\sqrt{\frac{G{m}_M}{r}}$

Where, $G$ is the gravitational constant, $r$ is the orbital radius, and $m_M$ is the mass of the Mercury.

Calculation:

It is known that the value of gravitational constant is

  $G=6.67\times {10}^{-11}\text{N}{\text{.m}}^{\text{2}}{\text{/kg}}^{\text{2}}$

The radius of the orbit of the satellite is

  $\begin{array}{c}r=R_M+h\\ =\left(2.44\times {10}^6\right)+\left(0.26\times {10}^6\right)\\ =2.7\times {10}^6\text{m}\end{array}$

Orbital speed can be calculated as follows:

  $\begin{array}{c}v=\sqrt{\frac{G{m}_M}{r}}\\ =\sqrt{\frac{6.67\times {10}^{-11}\times 3.30\times {10}^{23}}{2.7\times {10}^6}}\\ =2.85\times {10}^3{\text{m/s}}^2\end{array}$

Conclusion:

Thus, the orbital speed of the satellite is $2.85\times {10}^3{\text{m/s}}^2$ .

b.

To determine

The orbital period of the satellite.

Answer to Problem 17PP

  $1.65\text{h}$

Explanation of Solution

Given:

The mass of Mercury is $3.30\times {10}^{23}\text{kg}$ .

The radius of Mercury is $2.44\times {10}^6\text{m}$ .

Height of satellite above mercury’s surface is $260\text{km}=0.26\times {10}^6\text{m}$

Formula used:

Period of a Planet orbiting the Mercury is given by,

  $T=2\pi \sqrt{\frac{r^3}{G{m}_M}}$

Where, $G$ is the gravitational constant, $r$ is the orbital radius, and $m_M$ is the mass of Mercury.

Calculation:

It is known that the value of gravitational constant is,

  $G=6.67\times {10}^{-11}\text{N}{\text{.m}}^{\text{2}}{\text{/kg}}^{\text{2}}$

And, the radius of the orbit of the satellite is,

  $\begin{array}{c}r=R_M+h\\ =\left(2.44\times {10}^6\right)+\left(0.26\times {10}^6\right)\\ =2.7\times {10}^6\text{m}\end{array}$

Then, using the given and known values, the orbital period of satellite will be

  $\begin{array}{c}T=2\pi \sqrt{\frac{r^3}{G{m}_M}}\\ =2\pi \sqrt{\frac{{\left(2.7\times {10}^6\right)}^3}{6.67\times {10}^{-11}\times 3.3\times {10}^{23}}}\\ =2\pi \left(0.945\times {10}^3\right)\\ =5.938\times {10}^3\text{s}\times \frac{1\text{}\text{hr}}{3600\text{s}}\\ \approx 1.65\text{h}\end{array}$

Conclusion:

Thus, the orbital period of the satellite around Mercury is approximately $1.65\text{h}$ .