Angular Displacement: Earth's Orbit 2 Days

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salaam
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Homework Statement



Earth's orbit around the sun is nearly circular. The period is 1 yr = 365.25 days. In an elapsed time of 2.0 days, what is Earth's angular displacement?

Homework Equations



Theta=Theta final - Theta inital

The Attempt at a Solution



I'm very confused and need help starting it!
 
on Phys.org
It stands to reason that if Earth travels 365 degrees (2Pi radians) in 365.25 days, then Earth travels ___ degrees in 2 days. You should be able to set up an algebraic equation for this one.
 
salaam said:

Homework Statement



Earth's orbit around the sun is nearly circular. The period is 1 yr = 365.25 days. In an elapsed time of 2.0 days, what is Earth's angular displacement?

Homework Equations



Theta=Theta final - Theta inital

The Attempt at a Solution



I'm very confused and need help starting it!

if period is the time taken for one revolution and it is an approximate circle. Then in 1 time period (365.25 days) how many radians does it travel?
 
(2 * pi radians) / 365.25 = 0.0172024238
 
salaam said:
(2 * pi radians) / 365.25 = 0.0172024238

good good

so in 1 day it rotates 2π/365.25 radians.

So in 2 days how much does it rotate?
 
salaam said:
0.0344048476 radians

so in 2 days it rotates 0.0344048476 radians, isn't this what the question asked for ? (you can convert it degrees depending on what the question wants)
 
YES! thank you so much for your help :)
 
Is there anyway you can also help me find out the change in the Earth's velocity? i know that velocity is change in distance over time. but it keeps telling me that i can' tput my answer in radians/seconds so i dontk now how to do it
 
salaam said:
Is there anyway you can also help me find out the change in the Earth's velocity? i know that velocity is change in distance over time. but it keeps telling me that i can' tput my answer in radians/seconds so i dontk now how to do it

Well you know 2 days it rotates 0.0344048476 radians

the angular velocity ω is defined as d/dt or

[tex]\omega = \frac{\theta_2 - \theta_1}{t} = \frac{change \ in \ angular \ displacement}{time}[/tex]


and you have the change in angular displacement is 0.0344048476 radians.

So in 2 days what is ω ?
 
.0172 radians/ days.. but its not asking for angular velocity so it doesn't want my answers in radians/ days or hours or seconds
 
salaam said:
.0172 radians/ days.. but its not asking for angular velocity so it doesn't want my answers in radians/ days or hours or seconds

Yes but we need ω to get v. Convert ω to radians/second.

Now what is the relationship between v, ω and r ? (r is the distance from the center of rotation -> the sun in this case)
 
i converted w to radians/ second and got. 1.99 x 10^-7. i have no idea what the relationship is
 
OHH so i would just do v= 6378 km ( 1.99 x 10^-7) and that would give me my answer in km/second?
 
salaam said:
OHH so i would just do v= 6378 km ( 1.99 x 10^-7) and that would give me my answer in km/second?

Well I don't know the distance between the sun and the Earth, but that is what you would do.

Also if that is wrong, try using r=radius of sun + distance between the sun and the Earth + radius of the earth. Not sure they meant to use the sun and Earth as point masses or how they normally are.
 
OHH so i just do v= 6378 km ( 1.99 x 10 ^-7) and my answer comes out in km/s?