Is there life on mars? Questions

Question 1

Show that, with a surface gravity of 3.7 N/kg and a mass of 6.4 x 10²³ kg, Mars has a radius of about 3400km
G = 6.67 * 10^-11 Nm²/kg²

Answer 1

g = (-) GM/r²
so r² = GM/g
= 6.67 x 10⁻¹¹ x 6.4 x 10²³ / 3.7
= 1.15 x 10¹³
r = 3397 km
r ≈ 3400 km

Question 2

Calculate the gravitational potential at the surface of Mars, taking the radius to be 3390 km
M = 6.4 x 10²³ kg

Answer 2

V = (-) GM/r
= 6.67 x 10⁻¹¹ x 6.4 x 10²³/(3390 x 10³)
= 1.259233…x 10⁷
= 1.26 x 10⁷ J/kg

Question 3

Calculate the gravitational potential energy needed to lift a 2500 kg Mars lander back into orbit at 280 km
G = 6.67 * 10^-11 Nm²/kg²
M = 6.4 x 10²³ kg

Answer 3

E = (-) GmM/r²
= 6.67 x 10⁻¹¹ x 6.4 x 10²³ x 2500/(280 x 10³)²
= 1.3612… x 10⁶
= 1.36 x 10⁶

Question 4

Calculate the kinetic energy needed to lift a 2500 kg Mars lander back into orbit at 280 km

Answer 4

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Question 5

A day on Mars is 24 hours 40 minutes. Show that the altitude of an areostationary orbit is about 17,000 km

Answer 5

r = (h + rₘ)
mv²/r = GmM/r²
r = GM/v²
v=s/t
s = 2rπ
v = 2rπ / (246060 + 40*60)
2r²π = GMt
r² = 6.03 x 10¹⁷
r = 7.77 x 10 ⁸
h = 7.77 x 10 ⁸ - 3390 x 10³
????????

Question 6

Assuming the image of Mars is 500 x 500 pixels at 24 bit per pixel:
Calculate the resolution of the image

Answer 6

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Question 7

Assuming the image of Mars is 500 x 500 pixels at 24 bit per pixel:
Calculate the amount of information in the image

Answer 7

500 x 500 x 24 = 6 Mbits

Question 8

Assuming the image of Mars is 500 x 500 pixels at 24 bit per pixel:
Calculate the number of alternative colours that the image can contain

Answer 8

2^24 = 16,777,216
= 16,800,000 alternatives
(??)

Question 9

Assuming the image of Mars is 500 x 500 pixels at 24 bit per pixel:
Calculate the data transfer rate needed to transmit the image in 2 1/2 minutes

Answer 9

500 x 500 x 24 = 6 Mbits
6,000,000/(2.5*60)
= 40,000 bits/s

Question 10

Assuming the image of Mars is 500 x 500 pixels at 24 bit per pixel:
Describe how each of the following image processing techniques could improve the image of Mars:
Very brightness
Very contrast
Reduce noise
Detect edges
False colour

Answer 10

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Question 11

The orbital radius of Mars’ orbit is 2.3×10¹¹ m and that of earth is 1.5×10¹¹ m:
Use Kepler’s 3rd law T² ∝ r³ to show that the time for Mars to orbit the Sun is about 687 Earth days

Answer 11

(2.3×10¹¹ / 1.5×10¹¹)³/² = T / 365
1.89869… x 365 = T
T = 693.02… days
T ≈ 687 days (??)

Question 12

The orbital radius of miles is orbit is 2.3×10¹¹ m and that of earth is 1.5×10¹¹ m:
Calculate i) The orbital velocity and ii)The centripetal acceleration due to Mars’s orbit around the Sun

Answer 12

v²= GM/r
v² = 6.67 x 10⁻¹¹ x 6.4 x 10²³ / (3390 x 10³)
v² = 12.6 x 10⁶
v = 3549 m/s = 3500 m/s (2sf)

a = v²/r = 12.6 x 10⁶ / (3390 x 10³)
= 3.71455…m/s²
= 3.7 m/s² (2sf)

Question 13

The orbital radius of miles is orbit is 2.3×10¹¹ m and that of earth is 1.5×10¹¹ m:
Calculate the maximum and minimum times for a radio signal to travel from Earth to Mars and back

Answer 13

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Question 14

The mean surface temperature on Mars is 210 K:
Show that the root mean square speed of a gas molecule of mass m is given by cᵣₘₛ = √(3kT/m)

Answer 14

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Question 15

The mean surface temperature on Mars is 210 K:
Calculate the root mean square speed of i) CO₂ molecules ii) N₂ molecules

Answer 15

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Question 16

The mean surface temperature on Mars is 210 K:
So that the escape velocity for a gas molecule at the surface of a planet is given by v = √(2GM/r)

Answer 16

0.5mv² = GmM/r
v² =2GM/r
v = √(2GM/r)

Question 17

The mean surface temperature on Mars is 210 K:
Calculate the escape velocity for Mars

Answer 17

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Question 18

Use the Boltzman factor to calculate the ratio of N₂ : CO₂ Molecules which have a enough kinetic energy to escape from Mars’s gravity at the mean surface temperature of 210 K. What can you conclude from your answer?

Answer 18

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Question 19

By considering the energy required to move a gas molecules to a height h above the surface of a planet use the Boltzmann factor to show that p = p₀ e⁽⁻ᵐᵍʰ/ᵏᵀ⁾ where p₀ is the surface pressure, p is the pressure at height h

Answer 19

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Question 20

Calculate the pressure at the top of Mars’s largest extinct volcano Olympus Mons which is at a height of 22 km. you can assume that Mars’s atmosphere is 100% CO2

Answer 20

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Question 21

The background radiation level at the surface of Mars is on average 80mSv per year although it can reach as high as 20mSv in one day during a solar proton event. A dose equivalent of 1mSv gives a probability of developing cancer of 3% in an individual.
Calculate the risk of developing cancer due to a single solar proton event

Answer 21

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Question 22

The background radiation level at the surface of Mars is on average 80mSv per year although it can reach as high as 20mSv in one day during a solar proton event. A dose equivalent of 1mSv gives a probability of developing cancer of 3% in an individual.
If a colony of 500 individuals were to be established on Mars calculate an estimate of how many would be expected to develop a radiation induced cancer after a 20 year period on Mars

Answer 22

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Question 23

The distance from Mars go its L1 point is 320 times the radius of Mars. Solar protons can have energies up to 10keV:
Calculate the speed of 10keV solar protons

Answer 23

cᵣₘₛ = √(3kT/m)
=√ (3 x 10 x 1000 x 1.6 x 10⁻¹⁹ / (1.673 x 10⁻²⁷) )
= 1.69384… x 10⁶ m/s
= 1.7 x 10⁶ m/s (2sf)

Question 24

The distance from Mars go its L1 point is 320 times the radius of Mars. Solar protons can have energies up to 10keV:
Calculate the maximum angle through which solar protons must be deflected at L1 in order to miss Mars

Answer 24

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Question 25

The distance from Mars go its L1 point is 320 times the radius of Mars. Solar protons can have energies up to 10keV:
Calculate the distance that a 10keV solar proton would have to travel in a 2.0 μT uniform magnetic field at L1 in order to be deflected by enough to miss Mars

Answer 25

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