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avatar+1364 

 

i know this is a lot , but if someone can even give me an explation where I can begin this that would be a lot of help but if you can go trhough it all with me that would be great. I am very confused.

 Sep 26, 2019
 #1
avatar+2417 
+2

Take the elapsed time between its launch date and its contact-loss date. Then multiply that by its velocity, (make sure units match).

 

 

To calculate the position of pioneer 10 as of Janurary 2009, you take the elapsed time between the launch date and Janurary 2009, then take that and multiply it by the velocity to get the distance. (make sure units match)

 

 

The same methods can be applied for the other questions.

 


As for the logarithmic scale, I believe you either have learned to do that in class or in the past posts have learned it from other user's posts.

 Sep 26, 2019
 #2
avatar+1364 
0

for the first part after I multiply the veloctiy to get the distance wouldnt I then add distance between Sun and Pioneer 10 to the 6 years the pioneer 10 has covered in distance?

jjennylove  Sep 26, 2019
 #3
avatar+104969 
+3

Pioneer 10 launch date  = 3/2/1972

 

Days between this date   and 1/22/2003    =   11,284 days  = 270,816 hrs = 974,937,600 seconds

 

Assuming that the rate of  12.24 km/s   is the constant rate ( which is probably not constant)....the distance that Pioneer 10 was  from the Sun on 1/22/2003  =

 

Time  * Rate  = Distance

 

974, 937,600 s  * 12.24 km / s    ≈    11,933,236,224 km

 

So  multiplying this by 1000  for the number of meters we get

 

11,933,236,224,000  m

 

cool cool cool

 Sep 26, 2019
edited by CPhill  Sep 26, 2019
edited by CPhill  Sep 26, 2019
 #6
avatar+1364 
+1

Starting here at part 1, just did the calculations myself to double check and it all looks correct! I had solved this a bit differently but I beilive I was doing it all incorrect for this first step. smiley

jjennylove  Sep 26, 2019
 #4
avatar+104969 
+3

Second part

 

Days between launch date and 1/1/2009  = 13455 days  = 1,162,512,000 seconds

 

So....the distance from the Sun to Pioneer 10  as of Jan 2009  ≈

 

Rate * Time  = Distance

 

12.24 km /s * 1,162,512,000 s  =   14.229.146,880 km

 

Jenny....just add three 0's to this to get the number of meters....!!!

 

 

 

cool cool cool

 Sep 26, 2019
edited by CPhill  Sep 26, 2019
 #7
avatar+1364 
+1

Everyting here makes sense! Way more sense then what I was thinking to solve this. I was making it way more complicated than it was.

 

I have addedd the three zero's as well! smiley

jjennylove  Sep 26, 2019
 #5
avatar+104969 
+3

Third Part

 

Aldebaran is about  65 light years from the Sun  (Wikipedia)

 

Number of meters in one light year ≈  9,460,730,472,580,800 m

 

So.....the number of approx. meters from the Sun to Aldebaran  ≈ 

 

9460730472580800  * 65 = 

 

614,947,480,717,752,000 m

 

 

cool cool cool

 Sep 26, 2019
 #8
avatar+1364 
0

for the last part that says to make the logarithmic scale, how wuld i go forth in doing that ?

 

Not asking you to create the whole graph for me but as it goes for x and y axis, I can try my best to figure out the rest. Thank yousmiley

 

I just wanna make sure I know what im doing

jjennylove  Sep 26, 2019
 #9
avatar+104969 
+1

I.m not sure about  the log graph....but  I might go about it this way:

 

We can create a semi-log graph with the years being on the x axis (linear) and the distances  (as base 10 logs) are on the y axis

 

 

Let 1972  be  year 0

Let  2003  be year 20

Let 2009  be year  26

 

The distance from the Sun to the Earth  (the launch site of Pioneer 10) in 1972  is 140,000,000,000  m

Expressing this  as 10 raised to a power wehave  that   log 140,000,000,000  = 11.14

So ....the first point on the graph would be   (0, 11.14)

 

In 2003, Pioneer 10 was  11,933,236,224,000 m  from the Sun

So.....  log 11,933,236,224,000 ≈  13.08

So....the point (20, 13.08)  would be on this graph

 

In 2009  Pioneer 10  was  14.229.146,880,000 m  from the Sun

So....log 14,229146,880,000  ≈  13.15

So....the point  (26, 13.15)   would be on this graph

 

One last  thng.....   the distance from the Sun to Aldebaran  = 614,947,480,717,752,000 m

So....Log 614,947,480,717,752,000   = 17.78

This will not have a point associated  with it.....so.....I'll leave it to your discretion as how you want to show this

 

You would need to label the x axis as "Years Since 1972"

Label the y axis as " Base 10 Log Distances "

 

Here's the graph  (you would need to draw a curve through these points.....I don't know how to do that in Desmos) 

 

https://www.desmos.com/calculator/rsuu7k8yj6

 

That's about the best I can do with this   !!!

 

 

cool cool cool

 Sep 26, 2019
edited by CPhill  Sep 26, 2019
 #10
avatar+104969 
+1

Last part....

 

Time in years for  Pioneer 10  to go from Earth to Aldebaran  =


Distance  from Earth to Aldebaran  = 

 

Distance from Sun to Aldebaran -  Distance from Sun to Earth  =

 

614,947,480,717,752,000 m  - 140,000,000,000  m   =

 

614947340717752000 m

 

 

So  the number of seconds that it will take to travel from Earth to Aldebaran  =

 

Distance / Rate   = Time

 

(The rate that Pioneer travels = 12.24 km/ s  =  12240m / s ) 

 

614947480717752000 m / 12240 (m/s)   ≈   50240795810274  sec

 

The number of seconds in a year  = (3600/ hr * 24hrs/day * 365 days)

 

To find the  time in years we have

 

Time in Seconds / Number of Seconds in one year  =

 

50240795810274 / (3600 * 24 *365)   =  1,593,125 years!!!!

 Sep 26, 2019
edited by CPhill  Sep 26, 2019
edited by CPhill  Sep 26, 2019
edited by CPhill  Sep 26, 2019

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