Value of g Calculation and Equation
Value of g in fps
The acceleration felt by a free-falling object due to the gravitational force of the mass body is called gravitational acceleration and is expressed by g calculated using SI unit m/s2. The value of g depends on the mass of the body and its size, and its value varies from body to body. The value of g on the moon is constant.
Acceleration Due to the Gravity of the Moon
The acceleration due to the gravity of the moon or the magnitude of g on the moon is 1,625 m/s2.
Calculate the acceleration due to the gravity of the moon
The acceleration due to the formula of gravity is indicated by
G = GM / R2
Where,
G is the universal gravitational constant, G = 6.674 x 10-11 m3 kg-1 s-2.
M is the mass of the body measured using kg.
R is the mass body radius measured by m.
g is the acceleration due to the gravity determined by m/s2.
The mass of the moon is 7.35 × 1022Kg.
The radius of the moon is 1.74×106m
Substituting the values in the formula we get-
g= 6.67×10−11 × 7.35 × 1022 / (1.74×106)2
Thus, the value of g on the moon is g=1.625m/s2.
The Acceleration Due to Gravity also Follows the Unit of Acceleration
Newton's Law of Gravitation as applied to the Earth is F = G m M / r2, where F is the gravitational force acting on the body of mass m, G is the universal gravitational constant, M is the mass of the Moon, and r is the distance of the body from the centre of the Sun. g is the factor in equation F = m g, so g is given as follows:
g = G M / r2
Both G and M are empiric constants, and g has an inverse-square relationship to r, the distance from the centre of the earth's mass.
There are two consequences of this:
Since the Earth is an ellipsoid, the distance from the centre of a point on the surface decreases with the latitude, increasing g.
The rotation of the planet creates an anti-gravity centrifugal effect, which is at the highest at the equator and zero at the poles.
These two effects are conspiring to generate a g rise in latitude. Their magnitudes are easily determined by simple geometry.
Altitude
The effect of latitude is calculated on the basis of the standard surface of the geoid, which is the spheroid at sea level. Points above sea level are progressively further away from the centre of the earth, so g decreases with altitude in a predictable manner.
The anomaly of Gravity.
In practice, the value of g varies somewhat from the geometrically predicted value to latitude and altitude. Positive variation is caused by the following:
the mass of the local above-sea-level topography
above-average density of underlying rocks
The components of the structure of the Earth have a variety of densities. The geometric model of gravity conceives the Earth as a collection of onion-skin layers, each with a uniform density (and this is almost the case). Each individual sheet, because of its uniform density, has its centre of mass corresponding to that of the Earth. Nonetheless, if the layer has a small patch of higher density material, the centre of the mass is shifted towards the patch, decreasing r, and thereby increasing g.
The Table Below Shows the Value of g at Various Locations from Earth's Centre.
As can be seen from both the equation and table above, the value of g varies inversely with the distance from the centre of the earth. In fact, the variation in g with distance follows an inverse square law, where g is inversely proportional to the distance from the centre of the earth. This inverse square equation means that, if the gap is doubled, the value of g decreases by a factor of 4. As the distance is tripled, the value of g is reduced by a factor of 9. And so on, too. This inverse square relationship is shown in the right-hand graphic.
The value of G that is gravitation is the most basic concept taught in science classes. Students learn about gravitation in earlier classes starting from sixth grade.
This article explains in-depth about the value of G inFPS, acceleration due to the gravity of the moon, the acceleration due to gravity also follows the unit of acceleration, lays out the value of G at various locations from Earth Centre. It is a very informative article and students who want to get a good score and want to get a clear understanding of the concept of the value of G should definitely read and learn this article in depth.
This article is curated by Vedantu’s expert team in order to simplify the concepts which may occur difficult to some students. This study material is available on Vedantu’s website. The link is easily accessible and it is available for free PDF download. Students can download the free PDF and learn in an offline environment which will help them to focus more on studies and will not clutter their brains with unnecessary information.
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More on Gravity
We all know that gravity is an invisible pulling force that pulls things towards the object’s centre. The gravitational force is studied as inherently linked to Mass. we all know that there is a gravitational force of attraction that is present in between every object. The gravitational force is proportional to the mass of the object and as the distance increases between them, the force weakens. Both the objects keep exerting an equal and attractive force on each other. A falling object attracts the earth with the same force as the earth attracts it.
The acceleration due to gravity on earth is also known as the value of g on earth is 9.8 m/s2. This indicates that if an object is falling freely, the velocity of that object will keep increasing by 9.8 every second.
The acceleration due to gravity on the moon, also known as the magnitude of g on the moon, is 1,625 m / s2. The formula of gravity is indicated by G = GM / R2. The mass of the moon is 7.35 × 1022Kg.
The radius of the moon is 1.74×106m
Substituting the values in the formula we get-
g= 6.67×10−11 × 7.35 × 1022 / (1.74×106)2
Thus, the value of g on the moon is g=1.625m/s2.
FAQs on Value of g
1. What is the standard value of acceleration due to gravity (g) on Earth?
The standard value for the acceleration due to gravity on the surface of the Earth is approximately 9.8 m/s² (metres per second squared). This value is an average and can vary slightly depending on location, altitude, and the Earth's local geology.
2. What is the formula used to calculate the value of 'g'?
The value of acceleration due to gravity ('g') for a celestial body like the Earth can be calculated using the formula derived from Newton's Law of Universal Gravitation:
g = GM / R²
Where:
- G is the Universal Gravitational Constant (approx. 6.674 × 10⁻¹¹ N·m²/kg²).
- M is the mass of the celestial body (e.g., Earth).
- R is the radius of the celestial body.
3. What does 'gravitational acceleration' (g) actually represent?
Gravitational acceleration, or the 'value of g', represents the rate at which the velocity of an object increases when it is in free fall under the influence of gravity alone, neglecting air resistance. For example, a 'g' value of 9.8 m/s² means that for every second an object falls, its downward speed increases by 9.8 metres per second.
4. How does the value of 'g' differ from the Universal Gravitational Constant (G)?
The value of 'g' and the Universal Gravitational Constant 'G' are fundamentally different concepts often confused by students. Here are the key differences:
- Definition: 'g' is the acceleration experienced by an object due to gravity, while 'G' is a constant of proportionality that defines the strength of the gravitational force between any two masses in the universe.
- Value: 'g' has a variable value (e.g., ~9.8 m/s² on Earth, ~1.6 m/s² on the Moon), whereas 'G' is a universal constant with a fixed value of approximately 6.674 × 10⁻¹¹ N·m²/kg².
- Dependency: The value of 'g' depends on the mass and radius of the planet or celestial body. The value of 'G' is constant everywhere in the universe.
5. Why is the value of 'g' maximum at the poles and minimum at the equator?
The value of 'g' varies across the Earth's surface for two primary reasons:
- Earth's Shape: The Earth is not a perfect sphere; it is an oblate spheroid, slightly flattened at the poles and bulging at the equator. This means the polar radius is less than the equatorial radius. Since 'g' is inversely proportional to the square of the radius (R²), the smaller radius at the poles results in a higher value of 'g'.
- Earth's Rotation: The rotation of the Earth creates a centrifugal force that acts outwards, opposing the force of gravity. This effect is strongest at the equator and zero at the poles. This outward force slightly reduces the net acceleration felt, making 'g' minimum at the equator.
6. How does the value of 'g' change with altitude and depth?
The value of 'g' decreases as you move away from the Earth's surface, both upwards and downwards.
- With Altitude: As you go up from the surface, your distance (R) from the Earth's centre increases. According to the formula g = GM/R², an increase in R causes the value of 'g' to decrease.
- With Depth: As you go down into the Earth, the mass of the Earth pulling you towards the centre decreases (as the shell of mass above you exerts no net gravitational pull). This reduction in effective mass (M) causes the value of 'g' to decrease, becoming zero at the very centre of the Earth.
7. How does the value of 'g' on the Moon compare to that on Earth, and why?
The value of 'g' on the Moon is approximately 1.62 m/s², which is about 1/6th of the value on Earth. This significant difference is due to the Moon's much smaller mass and radius. The formula g = GM/R² shows that acceleration due to gravity is directly proportional to the body's mass and inversely proportional to the square of its radius. The Moon's smaller mass is the dominant factor causing its weaker gravitational pull.
8. Does the mass of a freely falling object affect the value of 'g' it experiences?
No, a common misconception is that heavier objects fall faster. In reality, the acceleration due to gravity ('g') is independent of the mass of the falling object. While a more massive object experiences a greater gravitational force (F=mg), it also has greater inertia (resistance to change in motion). These two effects perfectly cancel each other out, causing all objects in a vacuum to fall with the same acceleration, 'g'.
