Acceleration Due to Gravity of the Sun

In summary, Saladsamurai states that the acceleration due to gravity of the sun at the distance of the Earth's orbit is 0.00593036 meters per second2.
  • #1
ideefixem
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?
 
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  • #2
Your distance term is not correct. You need to convert it to meters before you square it.
 
  • #3
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
How about this...

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

= 0.005899244 m/s^2 ~0.00
 
  • #5
ideefixem said:
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
I'm using the distance from the Earth to the Sun.
1.50 x 10^11 = r
r^2 = 2.25*10^22
 
  • #7
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
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
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
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
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
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.
 

1. What is the acceleration due to gravity of the Sun?

The acceleration due to gravity of the Sun is approximately 274 meters per second squared. This means that for every second an object falls towards the Sun, its speed increases by 274 meters per second.

2. How does the acceleration due to gravity of the Sun compare to Earth's?

The acceleration due to gravity of the Sun is much greater than Earth's, which is approximately 9.8 meters per second squared. This is because the Sun has a much larger mass and therefore exerts a stronger gravitational pull on objects.

3. Does the acceleration due to gravity of the Sun vary based on distance?

Yes, the acceleration due to gravity of the Sun does vary based on distance. According to the inverse square law, the force of gravity decreases as the distance between two objects increases. This means that the acceleration due to gravity of the Sun decreases as the distance between an object and the Sun increases.

4. How does the acceleration due to gravity of the Sun affect the orbits of planets?

The acceleration due to gravity of the Sun plays a crucial role in the orbits of planets. It is responsible for keeping planets in their elliptical orbits around the Sun and also determines the speed at which they move.

5. Can the acceleration due to gravity of the Sun be measured?

Yes, the acceleration due to gravity of the Sun can be measured using Newton's second law of motion (F=ma). By measuring the force of gravity between the Sun and an object and the mass of the object, the acceleration due to gravity of the Sun can be calculated.

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