I"ve seen numbers such as 274 m/s2, but that doesn"t make sense to me seeing that relatively weak sources of satilikozelucak.comcraft 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.


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Raising an orbit with a weak form of propulsion

A satilikozelucak.comcraft 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 satilikozelucak.comcraft 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 satilikozelucak.comcraft points the engine behind itself and pushes itself forward. This causes the satilikozelucak.comcraft 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 satilikozelucak.com (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 satilikozelucak.com 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 9.7983Sun 695,700,000 1.3271E+20 274.20

Conclusion

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).