What is the gravitational acceleration of the Sun?

What is the gravitational acceleration of the Sun?

I've seen numbers such as 274 m/s2, but that doesn't make sense to me seeing that relatively weak sources of spacecraft thrust such as Electric Propulsion can overcome the Sun's gravity and raise heliocentric orbits as well as orbits around the Earth which has a gravitational constant of only 9.81 m/s2.

• that's not the gravitational constant – JCRM Nov 15 '19 at 21:49
• I've made an edit to the question to adjust some terminology. I think I can understand what the OP is getting at, so I don't see any need to shut the question down. I've also posted an answer. voting to leave open. – uhoh Nov 16 '19 at 0:31
• ...and now voting to reopen! – uhoh Nov 16 '19 at 11:19

Raising an orbit with a weak form of propulsion

Yes the Sun's gravity is stronger than the Earth's on each body's surface, but it drops like $$1/r^2$$. See the math below.

A spacecraft in a heliocentric orbit around the Sun will just continue to orbit the Sun without any propulsion for millions or possibly billions of years because the spacecraft is launched from Earth and will have the Earth's roughly 30 km/s velocity.

If you have a weak form of propulsion, it doesn't need to fight the Sun's gravity. Instead, the spacecraft points the engine behind itself and pushes itself forward. This causes the spacecraft to slowly spiral outwards over time.

Vocabulary:

Gravitational Constant is referred to as $$G$$. There's only one, and its value is 6.67430(15) × 10-11 m3 kg-1 s-2. The (15) is the one standard deviation uncertainty of the last two digits of 6.67430, so that's about 22 parts per million 1σ uncertainty.

At first that might seem huge, but the problem is that gravity is a pretty small force. For objects on Earth that we can measure the mass accurately, the gravitational force is so small that we can't measure it well. But for large objects in space (Earth, Moon, planets) where we can measure the gravitational force accurately by carefully measuring the motion of satellites, we have no way to independently determine the mass accurately.

So for accurate calculations in space we use the product $$G$$ times $$M$$ which is written $$GM$$ and called the standard gravitational parameter of an object. See for example Where to find the best values for standard gravitational parameters of solar system bodies? where the value for the Earth has 12 significant digits and the Sun's has 15! Compare that to only five significant digits for $$G$$ alone.

There is also a value called Standard gravity which is roughly the gravitational acceleration we experience on Earth. It is also called the standard acceleration due to gravity or standard acceleration of free fall. The numerical value is fixed and not measured, and defined as exactly 9.80665 m s-2 and written as $$g_0$$.

The acceleration you experience now is just written as $$g$$ with out the subscript.

Some math:

The gravitational acceleration at some distance $$r$$ from a point source of gravity or any spherically symmetric object (see Newton's Shell theorem) is given by

$$a = \frac{GM}{r^2}$$

Written in vector form it's

$$\mathbf{a} = -\mathbf{r} \frac{GM}{|r|^3} = -\mathbf{\hat{r}} \frac{GM}{|r|^2}.$$

$$r$$ is the position vector from the center of the body, so the minus sign tells you that the acceleration is downward.

The total acceleration you would experience on the Sun's or Earth's surface is complicated by the fact that no realistic body has a perfectly spherical mass distribution and most bodies are also rotating and other nearby bodies also pull on you. See for example @DavidHammen's interesting table of those for someone on Earth and this answer for an equation for the main term of acceleration due to Earth's oblateness.

The standard gravitational parameters and radii for the Earth and Sun are shown below, along with the approximate acceleration due to gravity on their surfaces using the equation above

Body      radius (m)      GM (m^3/s^2)    g (m/s^2)
------   ------------     ------------    ---------
Earth      6,378,137       3.9860E+14       9.7983
Sun      695,700,000       1.3271E+20     274.20


Conclusion

These values for $$g$$ at the surfaces are the same as the ones in your question.

They are equal to $$G$$ times $$M$$ divided by $$r^2$$ and are approximately the gravitational acceleration you would feel on each body's surface. They are not exact because the bodies are not spherically symmetric, and you would "feel" other accelerations both real (due to gravitational attraction mostly from the Moon on Earth and mostly from Jupiter on the Sun) and fictitious (centrifugal force due to the body's rotation).

Gravitational acceleration is not just one number for a given body

Gravitational acceleration for any body is a function of the body's mass and the distance from the body's center of mass at which you are measuring it. It is proportional to mass and inversely proportional to the square of the distance; double the distance and acceleration divides by 4. The 274 m/s2 value occurs at the Sun's surface (a somewhat ambiguous point, as I commented). The Earth's orbit has a radius roughly 200 times that of the Sun's surface, so the Sun's gravitational acceleration is some 200 x 200 times weaker out here than at its surface; on the order of a few cm/s2.

When you are already in orbit, you do not have to overcome gravity

All objects in orbit around the Earth are already also in orbit around the Sun, because the Earth is in orbit around the Sun. A spacecraft which acquires enough velocity to escape Earth won't necessarily escape the Sun, but won't fall into it either. Once you are in orbit, changing velocity will affect the trajectory of your orbit; any change in velocity, no matter how small, will have some effect, no matter how small, on your orbital trajectory. So you can apply even the tiniest force, but if you keep doing it long enough, the effect will accumulate. So, even a tiny thruster, if it keeps working for long enough, can take a spacecraft out of Earth's orbit, out of the Sun's orbit, and if it could continue to work long enough, out of the orbit we are in around the galactic center.

Once you're already in orbit you can increase your velocity and semi major axis even if your acceleration is a small fraction of the local gravity field. But long, gradual burns give you a spiral trajectory rather than elliptical transfer orbits. See Adler's answer to my question General guidelines for modeling a low thrust ion spiral?

As for the sun's gravity, yes it is huge near the sun. But strength of gravity falls with inverse square of distance. Double the distance and it's 1/4 as strong. Triple the distance and it's 1/9 as strong. Ten times the distance gives a field 1/100 as strong.

Sun's gravity in the neighborhood of earth is 6 millimeters/sec^2. Out in the Main Asteroid Belt it is less than 1 millimeter/sec^2. So for heliocentric orbits a little further out, ion might be adequate for an impulsive burn injection into an elliptical transfer orbit.