A satellite is held in orbit by a 2000- N gravitational
1 answer:
Answer:
The work done by gravity is zero.
Explanation:
Given that,
Gravitational force = 2000 N
Circumference = 80000 km
We need to calculate the work done by gravity
Using formula of work done
Here, = 0
Because, for a circular motion is always perpendicular to the force.
Where, F = force
d = distance
Put the value into the formula
Hence, The work done by gravity is zero.
You might be interested in
Answer:
The initial velocity should be greater than 23.9 m/s. This velocity is underestimated because it does not take into account the resistance of the air.
Explanation:
Hi there!
The final position vector of the car (denoted as "r" in the figure) is
r = ( 120, -100) m
(placing the origin of the frame of reference at the launching point).
The position vector at a time "t" is calculated by the following equation:
r = (x0 + v0 · t · cos θ, y0 + v0 · t · sin θ + 1/2 · g · t²)
Where:
x0 = initial vertical position.
v0 = initial velocity.
t = time.
θ = launching angle.
y0 = initial vertical position.
g = acceleration due to gravity.
At final time:
120 m = x0 + v0 · t · cos θ
-100 m = y0 + v0 · t · sin θ + 1/2 · g · t²
Since the origin is placed at the launching point, x0 and y0 = 0, then:
120 m = v0 · t · cos θ
-100 m = v0 · t · sin θ + 1/2 · g · t²
We have a system with two equations and two unknowns, so, we can solve it.
Solving the first equation for "v0":
120 m = v0 · t · cos θ
v0 = 120 m / (cos θ · t)
Replacing v0 in the second equation:
-100 m = v0 · t · sin θ + 1/2 · g · t²
-100 m = (120 m/cos 15° · t) · t · sin 15° - 1/2 · (9.8 m/s²) · t²
-100 m = 120 m · tan 15° - 4.9 m/s² · t²
(-100 m - 120 m · tan 15°) / (-4.9 m/s²) = t²
t = 5.2 s
Now we can calculate the initial velocity:
v0 = 120 m / cos 15° · 5.2 s
v0 = 23.9 m/s
The initial velocity should be greater than 23.9 m/s
This velocity is underestimated because it does not take into account the resistance of the air.
