What is the value of g at one earth radius?
Table of Contents
- 1 What is the value of g at one earth radius?
- 2 What will be the weight of an object on earth if the earth shrinks to the the half of its radius and its mass remains the same *?
- 3 What would be the value of g on the surface of the earth if it’s mass was twice and its radius half of what it is now?
- 4 What will be the value of G at the Centre of the Earth?
- 5 What will be the value of G at the Centre of the earth?
- 6 How do you calculate the value of g on the surface of the earth?
- 7 What is the acceleration due to gravity when the Earth shrinks?
- 8 What is the radius and acceleration due to gravity at its surface?
- 9 How do you calculate the mass of a planet with lost radius?
What is the value of g at one earth radius?
9.8 m/s2
If the value 6.38×106 m (a typical earth radius value) is used for the distance from Earth’s center, then g will be calculated to be 9.8 m/s2….The Value of g Depends on Location.
| Location | Distance from Earth’s center (m) | Value of g (m/s2) |
|---|---|---|
| Earth’s surface | 6.38 x 106 m | 9.8 |
| 1000 km above surface | 7.38 x 106 m | 7.33 |
What will be the weight of an object on earth if the earth shrinks to the the half of its radius and its mass remains the same *?
If the earth shrinks of half of its radius its mass remaining same the weight of the object on the earth will change to. ∴F1=4F2.
What would be the value of g on the surface of the earth if it’s mass was twice and its radius half of what it is now?
What would be the value of g on the surface of the earth if its mass was twice and its radius half of what it is now? g2=8g1 Thus , the value of g on the surface of the earth would be eight times the present value.
How the rotation of Earth affect the value of G?
As we move from the equator to the poles, the distance of the point of surface of earth from the centre of earth decreases. Hence, the acceleration due to gravity increases. Variation in acceleration due to gravity will be accordingly changed.
Why value of g is zero at the Centre of the earth?
At the centre of Earth, force due to any portion of the Earth at the centre will be cancelled due to the portion opposite to it. Thus, the gravitational force at the centre on any body will be 0. Since, mass m of an object can never be 0. Therefore, when F = 0, then g has to be 0.
What will be the value of G at the Centre of the Earth?
0
g=0 at the centre of Earth.
What will be the value of G at the Centre of the earth?
How do you calculate the value of g on the surface of the earth?
Answer: use the formula: force of gravity = mg, where m is the mass of the object andg is the acceleration of the object due togravity. Since g is always 9.8 m/s^2, just multiply the object’s mass by 9.8 and you’ll get its force of gravity!
Will the value of g be the same everywhere on the surface of the earth?
The value of g is not same everywhere on surface of earth because of the altitude or the height of the surface above the sea level is not same everywhere.
How does the radius of Earth affect the value of G?
Since the main deviation is actually to reduce the measured value of g, you will see g increase by an additional small increment over the 2\%. The value of “g” does not depend upon the amount of matter, but depends on the amount of mass. So changing the radius of Earth will not affect “g” as long as you don’t change its mass.
What is the acceleration due to gravity when the Earth shrinks?
When the earth shrinks 1.5 \%, the radius becomes R′ = R−R(1.5/100) = (9.85/100)R or R′/R = 98.5/100 After shrinks, the acceleration due to gravity becomes, g′ = R′2GM
What is the radius and acceleration due to gravity at its surface?
The radius of the earth is R and acceleration due to gravity at its surface is ‘ g ‘. If a body of mass ‘ m ‘ is sent to a height of R/4 from the earth’s surface, the potential energy is:
How do you calculate the mass of a planet with lost radius?
The way to approach this question is to use the gravitational equation and the definition of g. g’ = GM/ (0.99R)^2 = g/0.99^2 ie… it ouwld be 0.98g or 98\% of the current surface gravity. If you lost mass with the lost radius, then you need to model the mass at the surface of the Earth.