Answer:
Q1. corect is B, Q2. it is A, Q3. E and Q4. A
Explanation:
Q1 For this exercise we can use Newton's second law where acceleration is centripetal.
F = m a
a = v² / r.
G m M / r² = m v² / r
G M / r = v²
The velocity has a constant magnitude whereby we can divide the length of the circular orbit (2π r) between the period
G M / r = (2π r / T)²
r³ = G M T2 / 4π²
Let's calculate
T = 103 day (24 h / 1 day) (3600 s / 1h) = 8,899 10⁶ s
r³ = 6.67 10⁻¹¹ 1.99 10³⁰ (8,899 10⁶) 2 / 4π²
r = ∛ (266.25 10³⁰)
r = 6.4 10¹⁰ m
The distance matches the value in part B
Q2 Astronauts have measured the acceleration of gravity, so we can use the second law with a body on the planet's surface
F = m g
G m M_p / R_p² = m g
G M_p / R_p² = g
M_p = g R_p² / G
They indicate that the radius of the planet is half the radius of the Earth
R_p = ½ R_earth
R_p = ½ 6.37 10⁶
R_p = 3.185 10⁶ m
Let's calculate
M_p = 8.2 (3,185 10⁶)² / 6.67 10⁻¹¹
M_p = 1.25 10²⁴ kg
The correct answer is A
Q3 We use Newton's second law again, with part Q1, where M is the mass of the planet and m is the mass of the moon
r³ = G M T² / 4π²
T = 63 days (24h / 1day) (3600s / 1h) = 5.443 10⁶ s
r³ = 6.67 10⁻¹¹ 1.25 10²⁴ (5.443 10⁶)² / 4π²
r = ∛ (62.56807 10²⁴)
r = 3.97 10⁸ m
The correct answer is E
Q4 To calculate this part let's use the conservation of mechanical energy,
Starting point The surface of the moon
Em₀ = K + U = ½ m v2 - G m M / r
Final point. Infinity with zero speed
= 0
Em₀ = Em_{f}
½ m v² - G m M / R = 0
v² = 2 G M / r
M = v2 r / 2G
r = 2 G M / v²
Since we don't know the radius of the moon, we will also use the equation in part 2
M = g r² / G
r = √ GM / g
Let's replace
2G M / v² = √ G M / g
4 G M / v⁴ = 1 / g
M = v⁴ / (g 4G)
M = 3000⁴ / (2.7 4 6.67 10-11)
M = 1.12 10²³ kg
corract is A
