Kepler's third law hypothesizes that for all the small bodies in orbit around the
same central body, the ratio of (orbital period squared) / (orbital radius cubed)
is the same number.
<u>Moon #1:</u> (1.262 days)² / (2.346 x 10^4 km)³
<u>Moon #2:</u> (orbital period)² / (9.378 x 10^3 km)³
If Kepler knew what he was talking about ... and Newton showed that he did ...
then these two fractions are equal, and may be written as a proportion.
Cross multiply the proportion:
(orbital period)² x (2.346 x 10^4)³ = (1.262 days)² x (9.378 x 10^3)³
Divide each side by (2.346 x 10^4)³:
(Orbital period)² = (1.262 days)² x (9.378 x 10^3 km)³ / (2.346 x 10^4 km)³
= 0.1017 day²
Orbital period = <u>0.319 Earth day</u> = about 7.6 hours.
Answer:
s₁ = 0.022 m
Explanation:
From the law of conservation of momentum:

where,
m₁ = mass of hockey player = 97 kg
m₂ = mass of puck = 0.15 kg
u₁ = u₂ = initial velocities of puck and player = 0 m/s
v₁ = velocity of player after collision = ?
v₂ = velocity of puck after hitting = 48 m/s
Therefore,

negative sign here shows the opposite direction.
Now, we calculate the time taken by puck to move 14.5 m:

Now, the distance covered by the player in this time will be:

<u>s₁ = 0.022 m</u>
To solve this problem, we must know the gravitational force
of the planet. The equation would be,

This would calculate the force between two objects with
masses m1 and m2 and the gravitational constant, G, is 6.67 x 10^-11 m3 s-2
kg-1 and with r as the distance between the objects.
Thus,
F = (6.67 x 10^-11 m3 s-2 kg-1) * (5.68 x 10^26 kg) * (65
kg) * ((1/6.03 x 10^7 m)^2)
F = 678 kg/s^2 or 678 N
Answer is letter B.
I think it would be that it has no atmosphere