<span>So we want to know why is there a difference between the force of gravity on the Moon and the force of gravity of the Earth. So the gravitational force between two objects depends on the masses of both objects. That can be seen from Newtons universal law of gravity. F=G*m1*m2*(1/r^2). So lets say we are holding an object of mass m=1kg on a height r=1m on the Moon and we are holding the same object on the Earth also on the same height of r=1m. The Gravitational force on the Earth will be Fg=G*M*m*(r^2) where M is the mass of the Earth. The force between the moon and that object will be Fg=G*n*m*(r^2), where n is the mass of the moon. Since mass of the Moon is much smaller than mass of the Earth, The gravitational force between the Moon and that body will be almost 6 times smaller than the gravitational force between the Earth and that body. So the correct answer is B. </span>
Air speed is how fast you are peddling over the ground, wind speed is how fast the wind is blowing above the ground.Depending on the correlation between the two with direction thrown the pilot knows what sort of resistance he is flying against.
Answer:
The direction angle θ of the resultant in the Polar (positive) specification is then θ = α + 60°. The Law of Cosines is used to calculate the magnitude (r) and the Law of Sines is used to calculate the angle (α).
Answer:

Explanation:
Given:
height above which the rock is thrown up, 
initial velocity of projection, 
let the gravity on the other planet be g'
The time taken by the rock to reach the top height on the exoplanet:
where:
final velocity at the top height = 0 
(-ve sign to indicate that acceleration acts opposite to the velocity)

The time taken by the rock to reach the top height on the earth:



Height reached by the rock above the point of throwing on the exoplanet:

where:
final velocity at the top height = 0 


Height reached by the rock above the point of throwing on the earth:



The time taken by the rock to fall from the highest point to the ground on the exoplanet:
(during falling it falls below the cliff)
here:
initial velocity= 0 



Similarly on earth:

Now the required time difference:

