Chapter 11, Problem 69Q

(a)

To determine

The closest distance from Mars so that it appears as a disk when seen from Earth, if the angular size of the naked eye is $60\text{ arcsec}$. Also, determine whether Mars gets this close to Earth or not.

Answer to Problem 69Q

Solution:

$2.34\times {10}^7\text{ km}$, Mars never comes this close to the Earth.

Explanation of Solution

Given data:

The angular resolution of the naked eye is $60\text{ arcsec}$.

Formula used:

The small-angle formula is written as:

$D=\frac{\alpha d}{206,265}$

Here, α is the angle subtended by the object (in arc seconds), d is the distance between the observer and the object, and D is the linear size of the object.

Explanation:

Refer to appendix 2. The diameter of Mars is:

$D=6794\text{ km}$

Determine the minimum distance for Mars to appear as a disk.

Use the small-angle formula, as shown below:

$D=\frac{\alpha d}{206,265}$

Substitute $60\text{ arcsec}$ for $\alpha$ and $6794\text{ km}$ for $D$.

$\begin{array}{c}6794\text{ km}=\frac{\left(60\text{ arcsec}\right)\times d}{206,265}\\ d=\frac{\left(6794\text{ km}\right)\left(206265\right)}{\left(60\text{ arcsec}\right)}\\ =2.34\times {10}^7\text{ km}\end{array}$

From table 11-3, the minimum distance between Mars and the Sun is $1.381\text{ au}$.

The distance between the Sun and Earth is $1\text{ au}$. Therefore, the minimum distance between Earth and Mars will be $1.381\text{ au}-1\text{ au}=0.381\text{ au}$.

$\begin{array}{c}{d}_{\min \text{ Mars to Earth}}=\left(0.381\text{ au}\right)\left(\frac{1.496\times {10}^8\text{ km}}{1\text{ au}}\right)\\ =57\times {10}^6\text{ km}\end{array}$

The distance between Mars and Earth must be $2.34\times {10}^7\text{ km}$ for Mars to appear as a disk. However, the distance between Mars and Earth is $57\times {10}^6\text{ km}$. So, it won’t appear as a disk.

Conclusion:

The distance of Earth from Mars for the latter to appear as a disk is $2.34\times {10}^7\text{ km}$. Now, the minimum distance between Mars and Earth is $57\times {10}^6\text{ km}$. So, Mars will never be this close to Earth.

(b)

To determine

Whether Earth will appear as a disk from Mars when seen by an astronaut or not. Also, explain whether that astronaut will be able to see the Earth and the Moon as separate objects or not.

Answer to Problem 69Q

Solution:

The Earth will not be visible to the astronaut as a disk. Also, due to high angular resolution, the Moon and Earth can be resolved easily other than the situation when they are in the same line with Mars.

Explanation of Solution

Given data:

The angular resolution of the naked eye is $60\text{ arcsec}$.

Formula used:

The small-angle formula is written as:

$\alpha =\frac{206,265D}{d}$

Here, α is the angle subtended by the object (in arc seconds), d is the distance between the observer and the object, and D is the linear size of the object.

Explanation:

The average separation between Earth and Moon is $3.84\times {10}^5\text{ km}$.

The distance between the Sun and Earth is $1\text{ au}$. Therefore, the minimum distance between Earth and Mars will be $1.381\text{ au}-1\text{ au}=0.381\text{ au}$.

$\begin{array}{c}{d}_{\min \text{ Mars to Earth}}=\left(0.381\text{ au}\right)\left(\frac{1.496\times {10}^8\text{ km}}{1\text{ au}}\right)\\ =57\times {10}^6\text{ km}\end{array}$

Recall the expression for the small-angle formula.

$\alpha =\frac{206,265D}{d}$

Substitute $3.84\times {10}^5\text{ km}$ for $D$ and $57\times {10}^6\text{ km}$ for $d$.

$\begin{array}{c}\alpha =\frac{206,265\left(3.84\times {10}^5\text{ km}\right)}{57\times {10}^6\text{ km}}\\ =1390\text{ arcsec}\end{array}$

The angular resolution is very high for the distance between the Earth and the Moon as seen from Mars. So, they can easily be distinguished. Also, the Earth-Moon system is viewed from another object in the solar system. So, the angular resolution will change. Further, there will be a certain situation, especially at the time when all three are in the same line, that is, the Moon is between Mars and Earth or Earth is between the Moon and Mars, when it will be too close that it can’t be resolved when viewed from the surface of Mars.

Conclusion:

The Earth will not be visible as a disk when seen from the surface of Mars. As the value of the angular resolution for the Earth-Moon system as seen from Mars is high, it can be resolved easily only when they are not in the same line, it will be difficult to resolve if Mars, Moon and Earth are in the same line.