5

Question 210 ptsThe Earth moves uniformly around the sun once every 365.25 days: The radius of the orbit is approximately 5 * 10^11 meters. Compute the ANGULAR acce...

Question

Question 210 ptsThe Earth moves uniformly around the sun once every 365.25 days: The radius of the orbit is approximately 5 * 10^11 meters. Compute the ANGULAR acceleration of the Earth (In meters/second/second) as it moves around the Sun. Do not enter your answer in sclentific notation;

Question 2 10 pts The Earth moves uniformly around the sun once every 365.25 days: The radius of the orbit is approximately 5 * 10^11 meters. Compute the ANGULAR acceleration of the Earth (In meters/second/second) as it moves around the Sun. Do not enter your answer in sclentific notation;



Answers

The radius of the earth's orbit around the sun (assumed to be circular) is $1.50 \times 10^{8} \mathrm{km},$ and the earth travels around this orbit in 365 days. (a) What is the magnitude of the orbital velocity of the earth, in $\mathrm{m} / \mathrm{s} ?$ (b) What is the radial acceleration of the earth toward the sun, in $\mathrm{m} / \mathrm{s}^{2} ?$ (c) Repeat parts (a) and (b) for the motion of the planet Mercury (orbit radius = $5.79 \times 10^{7} \mathrm{km}$ , orbital period $=88.0$ days $) .$

Okay, So this problem, we're given the radius of the earth, which is one point five zero times Turn the eighth killed kilometers and we know that it takes three and sixty five days for the earth rotates around the sun. So the time t here are the period is going to be three and sixty five days. Okay, so now we're being asked what is the vantage of the orbital velocity of the earth and meters for a second? So wanna find V the velocity and then in part B, it's asking, what is the radio acceleration of the earth toward the sun? Okay, so on the radio acceleration okay. And we'll work on port cnd after that. Okay, so we want thie overto velocity. So we know that velocity is just a distance which we call X over time. No. And we know the distance that the earth is going to travel. There's going to be two pi times the radius because it's just going to travel on in a circle. So we take the circumference of the circle that's gonna give the distance. So X is going to be two pi r. Okay. And now we have to confer our auntie to the proper units. So we went our to be in meters so we can convert the kilometers two meters then. So in order to convert kilometers two meters, we just multiply this value by thousand. Use another one thousand meters and one kilometer. So our radius is going to be one point five zero times ten to the eleventh meters. Okay, another time when the time to be for the period s o use period in time interchangeably here some of the same thing. So we were at the time to be in seconds, then sort of convert days. Two seconds. Okay. So basically, if we convert these days in two seconds here, he's in the conversion factor. Between second and days, we're going to get three point one six times ten and negative, Said Thompson of the seven seconds. Okay, I'm just going to ring three and six, five days into amount of seconds that that takes okay, so we have that now. They're in the proper units so we can plug in the value for our here plugging the value for tea and the blast equation, and then saw for their velocity. So If I do that, I'm going to get two point nine nine times ten to the fourth meters per second. That's them in the order of velocity and now compute the rate of acceleration. So we know the rate of acceleration is going to be these squared over R. So I plugged in the value for me here. I plug in the r value that I'm given when they sure ours and meters, though. So if I plug in those two values, I'm going to get five point nine five times to the neck of third meters per second squared. So give me the radio velocity. Okay, so what about parts C and D? So for seeing veer and saying we put repeat parts and B, but we're using the planet Mercury now. So mercury as Noble Preeti over Ordell radius of five point seven nine times to the seven kilometers and its orbital period is going to be eighty eight days. Okay, Obviously we're changing the our value and the period here. So see, you're going to find the velocity. So when you find the velocity when our is equal to five point seven nine times seven seven kilometers and the period tee or the time is going to be eighty eight point zero days. Okay, So I would fund the with those conditions then and then party is asking What is the radio acceleration when are in the period or these values? Okay, so it's put diddle marks here. Okay, So what we do to solve this is we just repeat the same exact process here, so use a different A different our value for mercury in a different period from mercury. Okay, but the exact processes, the exact same thing. We just have different values for our and for the period. Okay, so repeat the same process. Only an orbital velocity. So the formula Blasi Fay use the values for Auntie is going to be for point seven eight times, ten to the fourth meters per second and the radio velocity, which again his he squared over R. So I take the value for V here. I plug it in the rate of loss equation, and I used the r value that I've written out here by plugging those values. I'm going to get three point nine five times ten ending of second meters per second squared. Okay, so it's gonna give me the orbital velocity and the rate of acceleration for mercury

So the radios off earth, which is moving around the sun in a circular orbit, is given. Art is equal to one point for you into 10 ways to power 11 meter per second. Right? So the next two think is given It is taking one year to complete one revolution around the sun. So time period is 3. 60 for you. These okay? Me if I could clearly So it is 3 65 bees. So on solving it, you can convert it in two seconds. So how we will convert it in two seconds. One day means 24 hours on toe. Convert our into second. We should multiply by 3600. So this is the time period in second. Now, in order to find the velocity, we can use the formula toe by our over the under the substitute the values here. So the value of pies 3.14 radius is 1.5 10 years to 11 over the value Off T is 3 65 into 24 in two. 32nd. So on solving it the value off we comes out to be 3.0 into 10 to the power four meter per second. So this is the answer for a part, no little soul, for be part in be part centripetal acceleration is asked. So we know centripetal exploration formula is we Squire are so now substitute the value 3.0 into generous to power for is square and the value off our is 1.5 into 10 days to power 11. So on solving the value off A comes out to be 6.0 into 10 days to power minus three meter per second squared. So this is the answer for be barred. Thank you.

For port. We are going to find the angular momentum of the daily rotation of the Earth that should be pulling the moment of inertia of the earth times, the angular velocity of the earth. We're gonna treat the earth of the perfect sphere. This would be 2/5 times the mass of the Earth times the radius of the earth squared, multiplied by the daily angler velocity of the Earth. And so the daily angular momentum would be equaling 2/5 times 6.0 times 10 to the 24th kilograms multiplied by 6.4 times 10 to the sixth meters. Quantity squared, multiplied by two pi radiance for every one day. And we're gonna multiply this by one day for every 86,400 seconds. And we find that here the daily rather the angular momentum of the daily rotation of the Earth. This is equaling 7.1 times 10 to the 33rd kilograms, meter squared per second. And this would be your answer for party for part B. We want the ah, essentially the yearly revolution about the sun. So we're going to treat the earth as a particle and we can say that the angular momentum of the yearly rotation about the sun would be equaling I omega severely. We're going to treat the, uh We're going to treat the earth as a particle, essentially, And so this would be equaling m the mass of the earth times the distance between essentially the sun and the earth. And then we're gonna multiply this by the yearly angular velocity. And so this would be equaling 6.0 times. 10 to the 24th kilograms multiplied by 1.5 times 10 to the 11th meters. Quantity squared, multiplied by two pi radiance every year or 365 days. And then we're gonna multiply this by one day for every 86,400 seconds, and we find that the momentum the angular momentum of the yearly rotation of the Earth aboutthe son would be equal to 2.7 times 10 to the 40th kilograms meter squared per second. This would be our answer to part B. That is the end of the solution. Thank you for

For number 28 were giving the radius of the orbit of the earth going around the sun. And we're to find the velocity of the earth going around the sun. So velocities is it's changing position per time. It goes the whole way around the circumference of the earth. So two pi r to pie are. And I was given my radius right by the time it takes one year for the earth to go around the sun. So that's 365 days, and I'm just gonna put my conversions in here. If I'm gonna change days, two hours. I know there's 24 hours in a day. Our worst a minute. They're 60 minutes in an hour and then minutes two seconds, 60 seconds in a minute. I do a mess. Get three point. Oh, I should probably two significant figures. So 3.0 times. 10 to the fourth meters per second. Part B. We're supposed to find the angular or who asked you? So that's it's change. Don't fade A for time, right? I want to be in radiance. I know that one time around is two pi radiant. So it's basically this without the are without the radius. So two pi over the time again, this whole conversion. So through 65 that I'm going to change hours, two minutes, two seconds. I get 2.0, times 10 to the negative Seventh radiance and then for port. See, I'm asked to find the centripetal acceleration. So centripetal acceleration picnic that linear velocity squared over the time. No, over the radius. So I just found when you were, lest he appear gonna square that divided by the radius And I was given the radius at the beginning. 1.5 times. 10 to the 11th. I get 0.6 times. No, that's it. Point over six. And they're being meters per second squared. I should probably keep two significant figures. So had a significant figure there.


Similar Solved Questions

5 answers
KAaenIntranen Mcen neocencebcnc axpooracn Biundun C4e7nincieasccetdentConatCroterComecenamt(ncmeaeLomcaccomdeieIncicaJaclc UZ0 Oteginlcnu? CicrhCTCcareth RaldtCdenioITTHARE'Mpulalijn Jandard Cenlen amecocndetoe Nntthi ChaeJutuateter tlncic4raDonulela Blarcrrd duvdoncomanumtrMtniIncieticlickIcUtDaote
KAa en Intranen Mcen neocencebc nc ax pooracn Biundun C4e7n incieasc cetdent Conat Croter Comecenamt (ncmeae Lomcac comdeie Incica Jaclc UZ0 Otegi nlcnu? Cicrh CTCcareth Raldt Cdenio ITTHARE 'Mpulalijn Jandard Cenlen ame cocndetoe Nntthi Chae Jutuateter tl ncic4ra Donulela Blarcrrd duvdon coman...
5 answers
Solve the system by using the addition method,6r"+4v3ryty=25There are infinltely many solutions0 0 6 (O,D)The solution set the empty set, (llDDThe solutlon set Is finite set;The solutlon set IsContinue
Solve the system by using the addition method, 6r"+4v 3ryty=25 There are infinltely many solutions 0 0 6 (O,D) The solution set the empty set, (ll DD The solutlon set Is finite set; The solutlon set Is Continue...
5 answers
Queston 6KaAt STP what volume of NHz will be formed when 18.0 L of Hz react?Nz(g) 3 Hzlg)2 NHz(g)0 18.0 LM14.0 LKl12.0 4K224L
Queston 6 Ka At STP what volume of NHz will be formed when 18.0 L of Hz react? Nz(g) 3 Hzlg) 2 NHz(g) 0 18.0 L M14.0 L Kl12.0 4 K224L...
5 answers
Solve the following differential equationsy = Voyx -y? +xy'
Solve the following differential equations y = Voyx -y? +xy'...
4 answers
6.52. Evaluate SS Vre+v dr dy; where R is the region 22+v? = a?
6.52. Evaluate SS Vre+v dr dy; where R is the region 22+v? = a?...
5 answers
A resistor has voltage of 120 volts and a resistance of 20 ohms_ What is the power consumed?
A resistor has voltage of 120 volts and a resistance of 20 ohms_ What is the power consumed?...
5 answers
When hydrogen is a part of aithas a slight positive charge
When hydrogen is a part of a ithas a slight positive charge...
5 answers
4. How do genetics play role in genetics? What is difference between natural selection and evolution? Mkot 0rf0nt Afknn DAl coma Mkat Ara
4. How do genetics play role in genetics? What is difference between natural selection and evolution? Mkot 0rf0nt Afknn DAl coma Mkat Ara...
5 answers
Question 41 + csc(t) Simplify to a single trig function with no fractions_ 1 + sin(t)Question Help: QvideoSubmit QuestionMacBook
Question 4 1 + csc(t) Simplify to a single trig function with no fractions_ 1 + sin(t) Question Help: Qvideo Submit Question MacBook...
5 answers
But - the solubility of CaCOs can be increased substantially by Calcium carbonate is an "insoluble salt". equilibrium equations and explain why adding acid will acidifying the solution; Write all the relevant increase the solubility of calcium carbonate:
But - the solubility of CaCOs can be increased substantially by Calcium carbonate is an "insoluble salt". equilibrium equations and explain why adding acid will acidifying the solution; Write all the relevant increase the solubility of calcium carbonate:...
5 answers
(5.00 Points) 2 Which one of the following is not true about peptide bonds?a) Peptide bond formation at room temperature in an aqueous solution occurs non-spontaneously: b) It occurs between the amine group of the first amino acid and the carboxyl group of the second one_ c) Proteases can be used to break the peptide bonds. d) Peptide bond formation is a dehydration reaction.
(5.00 Points) 2 Which one of the following is not true about peptide bonds? a) Peptide bond formation at room temperature in an aqueous solution occurs non-spontaneously: b) It occurs between the amine group of the first amino acid and the carboxyl group of the second one_ c) Proteases can be used t...
3 answers
3. Solve the recurrence relation with the given initial conditions to find closed formula_ 1Oak-1 2104-2 2k , %0 1,0,
3. Solve the recurrence relation with the given initial conditions to find closed formula_ 1Oak-1 2104-2 2k , %0 1,0,...
5 answers
1 4 VnHe 1 1 W # 8 1 7 0 0 Il } 1 1 1 1
1 4 VnHe 1 1 W # 8 1 7 0 0 Il } 1 1 1 1...
5 answers
Assume that A and B are events. IfP(A∪B)=0.7,P(A)=0.3,andP(B)=0.65,findP(A∩B).
Assume that A and B are events. If P(A∪B)=0.7, P(A)=0.3, and P(B)=0.65, find P(A∩B)....
5 answers
Luslulwheto6eR. Then 6 +AL eigenvalue of A6 + 2a(L0 I 0liud Unt ulicr cleHvulllc O[(42 If0 # duermingorthogonally riagonalizer wher det(P| =1 which the qquadratic nencinted with(4.31 tind allvnlues delinite.positive(4.4) If aasketeh the cune representer byTAT=12Wa tYoulr :keleh Iust Mccn=iMterewpls (ll tle rOlalel axtswl? 0 Fotaton
Luslul wheto 6eR. Then 6 + AL eigenvalue of A 6 + 2a (L0 I 0 liud Unt ulicr cleHvulllc O[ (42 If0 # duerming orthogonally riagonalizer wher det(P| =1 which the qquadratic nencinted with (4.31 tind allvnlues delinite. positive (4.4) If a a sketeh the cune representer by TAT=12 Wa t Youlr :keleh Iust ...

-- 0.057931--