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.
The study material on the value of G focuses mainly on the exact topic and doesn’t include any sort of unnecessary information that is not needed in the study of this topic.
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. Value of g is… ?
The expression for acceleration due to gravity is given by g= GM = R2
as we know the radius of the earth at poles is minimum, so acceleration due to gravity will be maximum at poles because g is inversely proportional to R2.
Also, the net acceleration due to gravity is affected (decreases) by centrifugal force due to rotation of the earth.
poles are less affected by rotation also their radius is small so poles have a maximum value of g.
2. How do you Calculate the Value of g?
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.
3. What is gravitation?
Gravitation can be defined as the universal force of attraction which is constantly acting between all matter. It is a natural Phenomenon through which all the things in the universe including planets, galaxies, stars, moon everything is attracted to one another. Earth's gravity gives weight to physical objects and the gravity of the moon is responsible for causing the tides of the oceans. There are various theories that are developed around gravity and are extremely important in the study of physics and science. The concept of gravity and the value of G is discussed in this article giving a comprehensive understanding of the topic and helping students to learn in an easier and efficient manner as the definitions given are extremely simple to understand.
4. Where can I find study material related to the value of g?
The study material related to the value of G is easily available on Vedanta‘s website. The link can be accessed easily through the internet and it is available for free PDF download. The PDF file is available so that students can download the file and learn and understand the concept anywhere and everywhere they want. The expert teachers at Vedantu have tailored the content in a way that makes learning easier for students.
5. How to calculate the value of g on the moon?
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.
6. Who first calculated the value of g?
71 years after Newton‘s death in 1798 the first direct measurement of gravitational attraction between two bodies in the laboratory was performed by Henry Cavendish. Cavendish had calculated the relationship between the angle of rotation to the amount of torsional force.
7. What is gravitational acceleration?
Gravitational acceleration can be defined as the acceleration of an object in free fall within a vacuum. It basically means that there is a steady gain in speed which is caused by gravitational force also known as gravitational acceleration.
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