Question

ASK YOUR TEACHER As an astronaut, you observe a small planet to be spherical. After landing on the planet, you set off, walking always straight ahead, and find yourself returning to your spacecraft from the opposite side after completing a lap of 25.7 km. You hold a hammer and a falcon feather at a height of 1.43 m, release them, and observe that they fall together to the surface in 30.0 s. Determine the mass of the planet.

Answer 1

Answer:

7.9 x  $10^1^4$ kg

Explanation:

Note: G  is the gravitational constant i.e 6.6743×10⁻¹¹ m³/kg⋅s²

First determine the radius of the planet:

circumference 'C' = 2πR

25.7 = 2πR

R = 25.7/2π

R = 4.09 km

Next is to find the acceleration due to gravity:

Δy = 1/2 gt²

1.43 = 1/2 g(30)²

g = 0.003177 m/s²

By using the universal law of gravitation:  

g = GM / R²

0.003177 = (6.67 x $10^-^1^1$)M / (4.09  x $10^3$)²

M = 7.9 x  $10^1^4$ kg

Thus,the mass of the given planet is  7.9 x  $10^1^4$ kg

Answer 2

Given Information:

circumference of planet = C = 25.7 km

height of release = h = 1.43 m

time = t = 30 s

Required Information:

Mass of planet = M = ?

Answer:

Mass of planet = M = 7.963×10¹⁴ kg

Explanation:

The mass of the planet can be found using Newton's law of gravitation,

g = GM/R²

M = gR²/G

Where g is the gravitational acceleration at the planet, R is the radius of the planet and G is the gravitational constant.

G = 6.6743×10⁻¹¹ m³/kg⋅s²

First we need to find R and g

The relation between circumference and radius is given by

C = 2πR

Where R is the radius of planet

25.7 = 2πR

R = 25.7/2π

R = 4.09 km

From the equations of motion we have,

h = ut + ½gt²

Where u is the initial speed which is zero, t is the time and h is the height of release.

Re-arrange the above equation for g

h = ½gt²

g = 2h/t²

g = (2*1.43)/(30)²

g = 0.003177 m/s²

Finally, the mass of the planet is

M = gR²/G

M = 0.003177*(4.09×10³)²/6.6743×10⁻¹¹

M = 7.963×10¹⁴ kg

Therefore, the mass of the given planet is 7.963×10¹⁴ kg