Acceleration Due to Gravity of the Sun

  • Thread starter ideefixem
  • Start date
  • #1
18
0

Homework Statement



What is the acceleration due to gravity of the sun at the distance of the earth's orbit?

Homework Equations



law of gravity = Gm/r^2
Sun's Mass: 1.99x10^30 kg,
earth-sun distance: 150x10^6 km

The Attempt at a Solution



((6.67*10^(-11)) * (1.99*10^30))/ (2.25*10^12)

= 58992444.4444

Does this appear correct?
 

Answers and Replies

  • #2
hage567
Homework Helper
1,509
2
Your distance term is not correct. You need to convert it to meters before you square it.
 
  • #3
3,003
2
Out of curiousity...why is r="2.25" where did that come from?
What are you adding to the mean Earth-Sun distance of 1 AU?

Casey
 
  • #4
18
0
How about this...

((6.67*10^(-11)) * (1.99*10^30))/ (2.25*10^22)

= 0.005899244 m/s^2 ~0.00
 
  • #5
3,003
2
How about this...

((6.67*10^(-11)) * (1.99*10^30))/ (2.25*10^22)

= 0.005899244 m/s^2 ~0.00
Looks much better. But again, what are you using for a radius?

Casey
 
  • #6
18
0
I'm using the distance from the Earth to the Sun.
1.50 x 10^11 = r
r^2 = 2.25*10^22
 
  • #7
3,003
2
Ah. I did not notice that you already squared r. Silly me. Anyway, I do not know what degree of accuracy you are looking for, but you may want to account for the fact that by Newton's Shell Theorem, r would be the distance from the center of one mass to the center of the other.

Casey
 
  • #8
1
0
g=Gm/r^2
For sun,
g=6.67*10^-11*1.989*10^30/(695000000)^2
=274.51m/s

Isn't it right?
 
  • #9
3
0
The earlier posts were almost correct the Sun’s acceleration on the Earth is -
2pi * Orbital Velocity in (meters/sec ) / Orbital Period (in seconds)
= 29785.513 * 6.28318531 / 31557600
= 0.00593036 m/s2
 
  • #10
3
0
NB. The Orbital Velocity squared * Radius is a constant for all the planets.
Eg. OV^2 * R(AU) = approx 887177000

For the Earth using OV = 29785.52 and R(AU) = 1 then K = 887177201
For Saturn using OV = 9644.8848 and R(AU) = 9.5371 then K = 887177310
 
  • #11
40
0
without an doubt, the correct answer to the question is from Saladsamurai
because distance from earth to sun is 1.50x10^8 km or 1.50x10^11m
 
  • #12
3
0
Yes, the distance between sun and Earth is approx 1.5E+11
Since g = OV^2/R then it still comes out at 0.005914512
Orbital Velocity squared = 887176784.7
Check it yourself.

For a new look at orbits and gravity check out Red Mangon on Facebook.
 

Related Threads on Acceleration Due to Gravity of the Sun

  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
2
Views
917
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
1K
Replies
2
Views
683
  • Last Post
Replies
1
Views
802
  • Last Post
Replies
6
Views
4K
Top