**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