Question

The most efficient way to send a spacecraft from the earth to

another planet is to use a Hohmann transfer orbit(Fig. P13.79). If the orbits of

the departure and destination planets are circular, the Hohmann transfer orbit

is an elliptical orbit whose perihelion and aphelion are tangent to the orbits of

the two planets. The rockets are fired briefly at the departure planet to put the spacecraft into the transfer orbit; the spacecraft then coasts until it reaches

the destination planet. The rockets are then fired again to put the spacecraft

into the same orbit about the sun as the destination planet. (a) For a flight from

earth to Mars, in what direction must the rockets be fired at the earth and at

Mars: in the direction of motion or opposite the direction of motion? What

about for a flight from Mars to the earth? (b) How long does a one-way trip

from the earth to Mars take, between the firings of the rockets? (c) To reach

Mars from the earth, the launch must be timed so that Mars will be at the right

spot when the spacecraft reaches Mars’s orbit around the sun. At launch, what

must the angle between a sun–Mars line and a sun–earth line be? Use

Appendix F.

Short answer

The rocket is fired in opposite direction.

The time taken for one way trip from Earth to Mars between the firings of rocket is,$t=259\mathrm{days}$.

The angle between a Sun-Mars line and Sun-Earth line is,$\theta =44.1^{\circ }$.

Step-by-step solution

Identification of given data

The orbit radius of Earth is,$r_E=1.50\times {10}^{11}m$ .

The orbit radius of Mars is,$r_M=2.28\times {10}^{11}\mathrm{m}$.

The semi major axis radius is, $a=\frac{1}{2}\left(r_E+r_M\right)$.

Concept of Kepler’s third law

As stated in Kepler's third law, the period's square is directly proportional to semi-major axis of orbit's sphere of radius.

The law can be given as,

$T=\frac{2{\mathrm{\pi a}}^{{\displaystyle \frac{3}{2}}}}{\sqrt{G{m}_s}}$… (i)

Determine the direction of rocket

The arithmetic average of orbit can be evaluated as,

$E=\frac{-G{m}_{s}m}{2r}$

The rockets are launched in the opposite direction of movement, that is, in the direction which increases the speed, in order to go from the Earth's circular orbit to the transfer orbit. Once in Mars orbit, the rockets must be launched in the opposite direction of motion in order to boost the energy.

Intervals between circular orbits may be found in the spacecraft's energy when it moves from one to another in the transfer zone.

Thus, the rockets are launched in the opposite direction of speed while returning from Mars to the Earth.

Determine the half time period

The semi-axis radius can be given by,

$$\begin{gathered} a=\frac{1}{2}\left(r_E+r_M\right) \\ a=\frac{1}{2}\left[\left(1.50\times {10}^{11}m\right)+\left(2.28\times {10}^{11}m\right)\right] \\ a=1.89\times {10}^{11}m \end{gathered}$$

Substitute above value in equation (i),

$$\begin{gathered} t=\frac{T}{2}=\frac{\mathrm{\pi }{\left(1.89\times {10}^{11}\mathrm{m}\right)}^{{\displaystyle \frac{3}{2}}}}{\sqrt{\left(6.673\times {10}^{-11}N{m}^2/k{g}^2\right)\left(1.99\times {10}^{30}kg\right)}} \\ t=2.24\times {10}^{7}s=259days \end{gathered}$$

Thus, the time taken for one way trip from Earth to Mars between the firings of rocket

is, 259 days which is more than $8\frac{1}{2}\mathrm{months}$.

Determine the angle between a Sun-Mars line and Sun-Earth line

The Mars will be in a 360-degree orbit at this period.

The angle between a Sun-Mars and Sun-Earth can be evaluated by,

$$\begin{gathered} \theta =\left({360}^{\circ }\right)\frac{\left(2.24\times {10}^{7}s\right)}{\left(687d\right)\left(86,400s/d\right)} \\ \theta =135.9^{\circ } \end{gathered}$$

And the spaceship does a 180-degree turn. So, the angle between a Sun-Mrs like and Sun-Earth line is,$44.1^{\circ }$.