What are the equations for solving planetary motion problems?

In summary, we are given a planet with a radius of 12km spinning at 520revs/s. We need to find the average speed of a point on the equator over 2.5 revolutions, the average acceleration on the star's circumference over 3/4 of a revolution, the distance covered by a point on the equator in 1 second, and the displacement. Using the equations v=2piRf, ac = ((2pi*R)*f)^2/R, and d=√(R*ac), we can solve for these values by considering velocity, acceleration, distance, and displacement as vectors and using trigonometry and the cosine law.
  • #1
Plutonium88
174
0

Homework Statement



a planet with radius of 12km spins at 520revs/s
find:
a) avg speed of a point on the planets equator over 2.5 of a revolution
b)find avg acceleration on stars circumfrance over 3/4 of a rev
c)find distance covered by point on the equator in 1 second
d) find displaement

f=520rev/s r=12000m

Homework Equations


fc=mv^2/r
T=1/f
v=2piR/T

The Attempt at a Solution



a) if i construct a vector triangle using v1 and v2 at the beginning of 3/2 of a rev and at the end, i can believe i can solve for vavg this way making a triangle where Vavg opposes the angle of 144 degrees...
v=2piR/T
where T=1/f

so i get v=2Pi*Rf

i considered V1and V2 = vvavg^2 = 2(2Pi*R*f)^2-2(2Pi*R)^2cos144

is this okay?

b)m*ac=mv^2/R
where ac = centripetal accell
and v = 2pi*R*F

ac = ((2pi*R)*f)^2/R

c)m*ac= (d/t)^2/R

if t=1s

d=√(R*ac)

d)... i think i may have solved displacement above.. perhaps i am confusing the distance and displacement.
 
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  • #2
a) avg speed of a point on the planets equator over 2.5 of a revolution

Here, avg speed is asked not avg velocity and in this case avg speed will be same as instantaneous speed.

b)find avg acceleration on stars circumfrance over 3/4 of a rev

avg acceleration is change in velocity over time.

ac = ((2pi*R)*f)^2/R

this is instantaneous acceleration, if all the units are in standard form. f should be in radians per second.

c)m*ac= (d/t)^2/R

if t=1s

d=√(R*ac)

seems great. :smile:

d)... i think i may have solved displacement above.. perhaps i am confusing the distance and displacement.

distance is the path traversed, that is the curve along which you move (the object, here point on the equator). displacement is the length of the straight line connecting initial and final position. moreover, displacement is a vector but distance is not.
 
  • #3
NihalSh said:
Here, avg speed is asked not avg velocity and in this case avg speed will be same as instantaneous speed.



avg acceleration is change in velocity over time.

so since its 1/3 of the distance

V=D/T => V (D/3)/T =(2PiR*f)/3

this is instantaneous acceleration said:
f[/I] should be in radians per second.

units are wanted in m/s^2

M*ac = V^2/R
ac = (((D/(3/4))^2/T)/R



distance is the path traversed said:
So what can i do to find the displacement ?
 
  • #4
so since its 1/3 of the distance

V=D/T => V (D/3)/T =(2PiR*f)/3

what is 1/3 of the distance??:confused:

anyways, its still not right.

M*ac = V^2/R

its not the correct expression. the expression is M*ac =M* V^2/R

But the acceleration in this equation is instantaneous. You require Average Acceleration.

Average Acceleration = ((final velocity)-(initial velocity))/(time interval)

Use vector addition/subtraction for calculating it.

So what can i do to find the displacement ?

check the image to be clear. the chord represents displacement and the arc represents distance.

calculate Δθ from the given data. Then a bit of trigonometry should give you the answer.
 

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  • #5
a) vavg = 2*pi**R*F

cause velocity is the same at every point in circular motion

b) ac = V^2(avg)/R

d)Δθ=ΔS/R

where S is my distance from C (S2-S1)/R = S2/R
 
  • #6
and then use cosine law with the angle.. Also was a type for the formula mac=mv^2/R , it was indicated in the beginning.
 
  • #7
Plutonium88 said:
a) vavg = 2*pi**R*F

cause velocity is the same at every point in circular motion

Yes exactly.:smile:

Plutonium88 said:
b) ac = V^2(avg)/R

this is not the correct expression. you need to use vectors to solve this.
Where would the velocity vector be pointing before and after 3/4th of revolution? What will be their magnitude?
draw it on a piece of paper, that should help
remember Average Acceleration = ((final velocity)-(initial velocity))/(time interval)

Subtract them vectorially and find time from the given info.
Plutonium88 said:
d)Δθ=ΔS/R

where S is my distance from C (S2-S1)/R = S2/R

what is C ?...center point of circle??
the formula is right. calculate the distance just like you previously did and that distance is ΔS.
Plutonium88 said:
and then use cosine law with the angle.. Also was a type for the formula mac=mv^2/R , it was indicated in the beginning.
yes, that should solve it for you.
 

1. What is planetary motion?

Planetary motion refers to the movement of planets around a central point, such as a star. This motion is caused by the gravitational pull between the planets and the central object.

2. What causes planetary motion?

Planetary motion is primarily caused by the force of gravity. This force is determined by the mass and distance between the planets and the central object. Other factors, such as the shape of the orbits and the presence of other celestial bodies, can also impact planetary motion.

3. How do scientists study planetary motion?

Scientists study planetary motion through various methods, including observation with telescopes, mathematical calculations and simulations, and spacecraft missions. These methods allow scientists to track the positions and movements of planets over time and make predictions about future movements.

4. What is Kepler's Laws of Planetary Motion?

Kepler's Laws of Planetary Motion are three laws developed by astronomer Johannes Kepler in the 17th century. These laws describe the motion of planets around the sun and are still used today to understand and predict planetary motion. The laws state that planets move in elliptical orbits with the sun at one focus, a planet's orbital speed varies depending on its distance from the sun, and the square of a planet's orbital period is proportional to the cube of its semi-major axis.

5. How does planetary motion impact our daily lives?

Planetary motion has a significant impact on our daily lives, as it affects the length of days, seasons, and tides. The position of the planets also plays a role in astrology and can influence cultural and societal beliefs. Additionally, understanding planetary motion is crucial for space exploration and navigation, as it allows us to accurately predict the movements of planets and spacecraft.

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