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The value of 'g' reduce to half of its value at surface of earth at a height 'h' , then :- a) h= R b) h= 2R c) h= ( root 2 + 1 ) R d) h= (root 2 -1 ) R ans. is (d) can anyone explain its solution?
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The value of 'g' reduce to half of its value at surface of earth at a ...
**Solution:**

To solve this problem, let's first understand the concept of acceleration due to gravity and its variation with height.

The acceleration due to gravity, denoted by 'g', is the force experienced by an object due to Earth's gravitational pull. It is defined as the gravitational force per unit mass.

At the surface of the Earth, the value of 'g' is approximately 9.8 m/s^2. However, as we move away from the surface and go to higher altitudes, the value of 'g' decreases.

Now, let's consider a point at height 'h' above the surface of the Earth.

**Using the concept of gravitational force:**

The gravitational force acting on an object at height 'h' is given by the formula:

F = (G * M * m) / (R + h)^2

Where:
- F is the gravitational force
- G is the gravitational constant
- M is the mass of the Earth
- m is the mass of the object
- R is the radius of the Earth
- h is the height above the surface of the Earth

**Relation between 'g' and 'F':**

The acceleration due to gravity, 'g', is defined as the gravitational force per unit mass. Therefore, we can write:

g = F / m

**Equating the values of 'g' at the surface and height 'h':**

At the surface of the Earth, the value of 'g' is g0 (let's assume).

So, g0 = (G * M) / R^2

At height 'h', the value of 'g' is g/2 (given).

So, (g/2) = (G * M) / (R + h)^2

**Simplifying the equation:**

(g/2) = (G * M) / (R^2 + 2Rh + h^2)

g = (2 * G * M) / (R^2 + 2Rh + h^2)

**Comparing the two equations:**

Comparing the equations g0 = (G * M) / R^2 and g = (2 * G * M) / (R^2 + 2Rh + h^2), we can equate the two expressions for 'g' to find the value of 'h'.

(g/2) = g0

(2 * G * M) / (R^2 + 2Rh + h^2) = (G * M) / R^2

Cross-multiplying and simplifying, we get:

2R^2 = R^2 + 2Rh + h^2

R^2 - 2Rh + h^2 = 0

(R - h)^2 = 0

(R - h) = 0

h = R

Therefore, the height 'h' is equal to the radius of the Earth (R).

Hence, the correct answer is option (a) h = R.
Community Answer
The value of 'g' reduce to half of its value at surface of earth at a ...
We know g=GM/R^2Where R =radius of the earth.At a hight h above the earth surface radius becomes R+hSo we can write gh =GM/ (R+h) ^2=>gh/g=(R/R+h)^2=>gh=g(R/R+h)^2,given gh=g/2=>g/2=g(R/R+h)^2=>(R/R+h)^2=1/2=>R+h/R=root2=>R+h=R root2=>h=(root 2-1)R
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The value of 'g' reduce to half of its value at surface of earth at a height 'h' , then :- a) h= R b) h= 2R c) h= ( root 2 + 1 ) R d) h= (root 2 -1 ) R ans. is (d) can anyone explain its solution?
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The value of 'g' reduce to half of its value at surface of earth at a height 'h' , then :- a) h= R b) h= 2R c) h= ( root 2 + 1 ) R d) h= (root 2 -1 ) R ans. is (d) can anyone explain its solution? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about The value of 'g' reduce to half of its value at surface of earth at a height 'h' , then :- a) h= R b) h= 2R c) h= ( root 2 + 1 ) R d) h= (root 2 -1 ) R ans. is (d) can anyone explain its solution? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The value of 'g' reduce to half of its value at surface of earth at a height 'h' , then :- a) h= R b) h= 2R c) h= ( root 2 + 1 ) R d) h= (root 2 -1 ) R ans. is (d) can anyone explain its solution?.
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