Question

Your spaceship lands on an unknown planet. to determine the characteristics of this planet, you drop a 1.50 kg wrench from 5.50 m above the ground and measure that it hits the ground 0.811 s later. you also do enough surveying to determine that the circumference of the planet is 6.28×104 km . part a what is the mass of the planet, in kilograms?

Answer 1

1. calculate the value of acceleration that objects gains in that period of time
•calculating acceleration
5.50 = 1/2at^2
5.50*2/t^2 = a
11.00/0.657 = a
16.74=a
now you got the acceleration
2. you have laws of gravitation for that

g = Gm/r^2
where g is the acceleration value
16.74 = 6.754*10^-11 × m/ 6.28*10^4
105.14*10^4 /6.754*10-11 = m
15.567*10^15 = m
that would be the mass of the planet ...

Answer 2

Answer:

The mass of the planet is $M=2.5*10^{25}kg$

Explanation:

We have a first part of the problem, which we resolve with kinematics, knowing that

$d=d_{0}+v_{0}t+\frac{1}{2}at^2$

where d is given (5.50m), d₀ is zero, v₀ is zero too (as the wrench starts falling from static position), t is given (0.811s), and a is what we want to know for the second part of the problem. We clear a

$a=\frac{2*5.5m}{(0.811s)^2}=16.72\frac{m}{s^2}$

Then for the second part, we use Newton's gravitational Law, where

$F_{g}=G\frac{mM}{r^2}$

m is the wrench mass, M is the planet mass, G is the gravity universal constant, and r is calculated from the given circumference (with the correct units) as

$2\pi r=6.28*10^7m\Leftrightarrow r=\frac{6.28*10^7m}{2\pi}=9994930.4m$

Finally, we have that

$F_{g}=G\frac{mM}{r^2}\Leftrightarrow ma=G\frac{mM}{r^2}\Leftrightarrow a=G\frac{M}{r^2}\Leftrightarrow M=\frac{ar^2}{G}$

Therefore, replacing with the data calculated, and the known value of G, we can calculate M of the planet

$M=\frac{16.72*9994930.4^2}{6.67428*10^{-11}}=2.5*10^{25}kg$