Answer:
m = 1.26*10²⁵ kg.
Explanation:
Assuming that the mass of the stone is much smaller than the mass of the planet, we can get the mass, applying the Universal Law of Gravitation to both masses, as follows:
Fg = G* ms* mp / rp²
Now, if we apply Newton's 2nd Law to the mass of the stone, we can get the gravitational acceleration, as follows:
Fg = ms*a = ms*g ⇒ g = G*mp / rp²
First of all, we need to get the value of g.
Assuming that this acceleration is constant, we can appy the kinematic equations to this situation.
We know that the stone is thrown upward with an initial velocity vo = 15 m/s.
At the highest point in the trajectory, just before of changing direction, the stone comes momentarily to a stop.
At this point, applying the definition of acceleration, we can write:
vf = vo -g*t ⇒ 0 = vo -gt ⇒ g = vo/t (1)
We have the total time since the stone was thrown upwards, not the one used for the upward trajectory.
It can be showed, using the expression for the displacement (which is the same in both directions) that the time used for going up, it's the same used to go down, so the time that we need to put in (1). is just the half of the total time.
So, replacing in (1) we get the value of g, as follows:
g = 15 m/s / 4.5 s = 3.33 m/s²
Now, we can replace this value in the equation that gives us g based in the Universal Law of Gravitation, as follows:
g=G*mp / rp² (2)
Before solving for mp, however, we need to get the value of the radius of the planet.
Assuming that it's a perfect sphere, we can get this value from the value of the circumference at the planet's equator:
rp = 2*π*rp / 2*π ⇒ rp = 1.0*10⁵ km / 2*π = 15,915 km.
With this value for rp, we can solve (2) for mp, as follows:
mp= g*rp² / G = 3.33 m/s² * (15,915 km)² / 6,67*10⁻¹¹ N.m²/kg²
mp = 1.26*10²⁵ kg.
, where L is the length and g is the gravitational constant, is the formula for a pendulum's period.