Question

You joined the SQU science fiction club which is making an animation video about time travel to the early solar system when the moon was closer to Earth. Your task is to arrive at an accurate description of the orbital motion of the moon at the early solar system when it was at 0.80 the current distance from Earth. You summarized your task in the following questions that you are addressing in order:Hint: Remember the application of Kepler laws in Ch 7. The orbital period and average distance of the moon from Earth at the present time are 27.3 days and 385,000 km, respectively. The diameter of the Moon is 3474 km and the diameter of the Sun is 1.39x106 km. The Earth-Sun distance is approximately 150 million km. Assume that the Earth-Sun distance and the geometry of the orbital planes are the same then (early time) and now. Would the moon appear from Earth larger or smaller at this early time of the solar system compared to now? * It would appear smaller It will appear the same because it has the same radius It

It would appear larger

What is the angular size of the moon at this early time of the solar system? *

0.80 degree

0.64 degree

0.86 degree

0.69 degree

0.74 degree

Is it possible to include a total solar eclipse scene, i.e. is it possible a total solar eclipses could happen during this time? *

No

Yes

What was the moon’s orbital period at the early solar system when it was orbiting Earth at 0.80 the current distance? Choose the closest answer *

14.3 days

12.7 days

16.0 days

17.7 days

19.5 days

You have a telescope like that of Galileo (with 15 mm aperture) and you know that the moon is about 385,000 km from Earth now and that you can use 500 nano meter for the wavelength of visible light.

What is the smallest surface feature you can observe on the moon? Choose the closest answer. *

15.55 km

7.78 km

6.67 km

9.33 km

11.66 km

Answer 1

a) At the early time distance between the Earth and the Moon,

r = 0.8*385000 km

= 308000 km

diameter of the moon, d = 3474 km

angular size of the moon, theta = d/r

= 3474/308000

= 0.01128 radians

= 0.01128*360/(2*pi)

= 0.64 degrees <<<<<<------------Answer

b) Yes

c) we know, T is proportional to r^(3/2)

so, T = (0.8)^(3/2)*27.3

= 19.5 days <<<<<<<<-------------Answer

d)
minimum angular size of an object on the moon that can be resoved, theta = 1.22*lamda/d

= 1.22*500*10^-9/(15*10^-3)

= 4.07*10^-5 rad

minimum size of the object that can be seen,

s = r*theta

= 385000*4.07*10^-5

= 15.55 km <<<<<<<<----------------Answer