Note that the simulation allows you to also display the force of the smaller moon
1 answer:
the force that the planet exerts on the moon is equal to the force that the moon exerts on the planet
Explanation:
In this problem we are analzying the gravitational force acting between a planet and its moon.
The magnitude of the gravitational attraction between two objects is given by
where :
is the gravitational constant
m1, m2 are the masses of the two objects
r is the separation between them
In this problem, we are considering a planet and its moon. According to Newton's third law of motion,
"When an object A exerts a force (action force) on an object B, then object B exerts an equal and opposite force (reaction force) on object A"
If we apply this law to this situation, this means that the force that the planet exerts on the moon is equal to the force that the moon exerts on the planet.
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Answer:
A) 16.6 m/s i -17.2 m/s j B) 23.9 m/s c) 46º below horizontal.
Explanation:
A) Once released, the football is not under the influence of any external force in the horizontal direction, so it continues moving at a constant speed equal to the initial velocity, i.e., 16.6 m/s.
If we choose the horizontal direction to be coincident with the x-axis, and make positive the direction towards the right (assuming that this was the direction along which the football was thrown), we can write the horizontal component of the veelocity vector, as follows:
vₓ = 16.6 m/s i
In the vertical direction, the football, once released, is in free fall, starting from rest.
So, we can find the vertical component of the velocity vector, at a given point in time, applying the definition of acceleration, as follows:
vy = a*t = -g*t = -9.81 m/s²*1.75 s = -17.2 m/s
Assuming that the upward direction is the positive for the y-axis (perpendicular to the chosen x-axis), we can write the vertical component of the velocity vector, at t=1.75 s, as follows:
vy = -17.2 m/s j
So, the velocity vector, in terms of the unit vectors i and j, can be written in this way:
v = 16.6 m/s i -17.2 m/s j
b) The magnitude of this vector can be found applying trigonometry, as the magnitude is the hypotenuse of a triangle with sides equal to vx and vy, as follows:
v = 23.9 m/s
c) The direction of the vector (below the horizontal) can be found as the angle which tangent is given by the quotient between vy and vx, as follows:
tg θ =
⇒ θ = tg⁻¹ (-1.036) = 46º below horizontal.