What is the value of acceleration due to gravity on the surface of the earth?
9.8 m/s2
Understanding Acceleration Due to Gravity on Earth
The acceleration due to gravity is a fundamental concept in physics that describes the acceleration experienced by an object due to the gravitational force exerted by a massive body like the Earth. When an object is in free fall near the Earth's surface, it accelerates towards the center of the Earth because of gravity.
The value of this acceleration due to gravity is not exactly the same everywhere on the Earth's surface due to factors like altitude, latitude, and local geological structures. However, a standard or average value is commonly used for calculations and general understanding.
Standard Value of Acceleration Due to Gravity
The standard value for the acceleration due to gravity on the surface of the Earth is internationally defined. This value represents the average acceleration that an object would experience in a vacuum near sea level at a latitude of approximately 45 degrees.
The commonly accepted and widely used standard value for the acceleration due to gravity is \(9.8 \text{ m/s}^2\).
Analyzing the Given Options
Let's examine the provided options to determine which one matches the standard value of acceleration due to gravity on the Earth's surface:
- Option 1: \(9.6 \text{ cm/s}^2\)
This value is given in cm/s2. Converting to m/s2, \(9.6 \text{ cm/s}^2 = 9.6 \times 10^{-2} \text{ m/s}^2 = 0.096 \text{ m/s}^2\). This is significantly smaller than the expected value. - Option 2: \(9.8 \text{ m/s}^2\)
This value is \(9.8\) and the units are m/s2. This matches the standard accepted value for acceleration due to gravity on the Earth's surface. - Option 3: \(10.8 \text{ m/s}^2\)
This value is \(10.8\) and the units are m/s2. While the units are correct, the numerical value is significantly higher than the standard \(9.8 \text{ m/s}^2\). - Option 4: \(9.8 \text{ cm/s}^2\)
This value is \(9.8\) but the units are cm/s2. Converting to m/s2, \(9.8 \text{ cm/s}^2 = 9.8 \times 10^{-2} \text{ m/s}^2 = 0.098 \text{ m/s}^2\). This is also significantly smaller than the expected value.
Comparing the options with the standard value, Option 2 correctly represents the acceleration due to gravity on the surface of the Earth with the appropriate units.
| Option | Value | Units | Value in m/s<sup>2</sup> | Matches Standard \(9.8 \text{ m/s}^2\)? |
|---|---|---|---|---|
| 1 | 9.6 | cm/s<sup>2</sup> | 0.096 | No |
| 2 | 9.8 | m/s<sup>2</sup> | 9.8 | Yes |
| 3 | 10.8 | m/s<sup>2</sup> | 10.8 | No |
| 4 | 9.8 | cm/s<sup>2</sup> | 0.098 | No |
Therefore, the value of acceleration due to gravity on the surface of the earth among the given options is \(9.8 \text{ m/s}^2\).
Revision Table: Key Concepts of Acceleration Due to Gravity
| Concept | Description | Standard Value (Earth Surface) |
|---|---|---|
| Acceleration Due to Gravity (g) | The acceleration experienced by an object in free fall near a celestial body's surface due to its gravity. | \(9.8 \text{ m/s}^2\) |
| Units of Acceleration | Metre per second squared (m/s<sup>2</sup>) in SI units, centimetre per second squared (cm/s<sup>2</sup>) in CGS units. | \(9.8 \text{ m/s}^2\) (SI) \(980 \text{ cm/s}^2\) (CGS) |
| Factors Affecting 'g' | Altitude (decreases with height), Latitude (increases slightly towards poles), Earth's Rotation, Local Density (mass distribution). | N/A (Value varies) |
Additional Information: Variation of Acceleration Due to Gravity
While \(9.8 \text{ m/s}^2\) is the standard average value, the actual acceleration due to gravity varies across the Earth's surface. Some reasons for this variation include:
- Altitude: As you move away from the Earth's center (increase in altitude), the gravitational force decreases, and thus the acceleration due to gravity decreases.
- Latitude: The Earth is not a perfect sphere; it bulges slightly at the equator and is flattened at the poles. Objects at the poles are slightly closer to the Earth's center than objects at the equator. Also, the centrifugal force due to Earth's rotation opposes gravity more strongly at the equator than at the poles. Both factors cause the acceleration due to gravity to be slightly higher at the poles (approx. \(9.83 \text{ m/s}^2\)) than at the equator (approx. \(9.78 \text{ m/s}^2\)).
- Local Geology: Variations in the density of the Earth's crust in different locations can cause small local variations in gravity.
For most standard physics problems, the value \(9.8 \text{ m/s}^2\) is used unless otherwise specified.
Similar Questions
The acceleration due to gravity on the Moon is (1/6) of that on the Earth. Hence, an object weighing 12 N on the Earth will weigh ________ on the Moon.
Fill in the blank with the most appropriate option.
The Universal Constant of Gravitation is ________.
Which of the following statements is/are INCORRECT?
A. The ratio of the force of gravitation between two masses, m1 and m2, kept at a distance R on the earth and on the moon is 1:1.
B. Nm2/kg2 is the SI unit of G.
C. The value of G depends on the distance between the bodies.
D. The value of G depends on the masses of the bodies.Calculate the work done by the force of gravity when a satellite moves in an orbit of radius 40,000 km around the earth.
A body has a weight W on the surface of Earth. What is its weight on a planet whose mass is 15 times that of Earth and a radius that is 4 times that of the earth?
The value of g on the moon is 1/6 th of the value of g on the earth. If a man can jump 1.5 m high on the earth, on the moon, he can jump up to a height of:
Consider a hypothetical planet whose mass and radius are both half that of Earth. If g is the acceleration due to gravity on the surface of Earth, the acceleration due to gravity on the planet will be:
A 5 kg object is raised through a height of 4 m. The Work done by the force of gravity acting on the object is (take g = 10 m/s 2):
Consider a planet whose mass and radius are both twice the mass and radius of Earth. The acceleration due to gravity on the surface of the planet is n times that on Earth. The value of n is:
Important Questions from Gravity
Which of the following law states that, "The force between two objects is directly proportional to the product of their masses?"
Which of the following statements about the movement of planets is true?
A. A planet's orbit is elliptical with the Sun at one of two focal points.
B. The orbit of a planet is circular with the sun in the center.
C. The orbit of a planet is elliptical with another planet in one of the two center-points.
D. The orbit of a planet is circular with another planet in the center.
If the mass of a person is 60 kg on the surface of earth then the same person’s mass on the surface of the moon will be:
The centripetal force required to keep the moon in its orbit is provided by which force?
How is the acceleration due to gravity denoted?
