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Chapter 8: Problem 16

The height of the point vertically above the earth's surface, at which acceleration due to gravity becomes \(1 \%\) of its value at the surface is (Radius of the earth \(=R\) ) (a) \(8 R\) (b) \(9 R\) (c) \(10 R\) (d) \(20 R\)

Short Answer

Expert verified
None of the options (a), (b), (c) and (d) are correct. The correct answer is that the height above the Earth鈥檚 surface at which acceleration due to gravity becomes 1% of its value at the surface is 99R, where R is the radius of the Earth.

Step by step solution

01

Understanding the question

We know that the acceleration due to gravity (\(g\)) at the surface of the Earth is \(9.8 m/s^2\). But at a height \(h\) above the surface of the Earth, the acceleration due to gravity becomes 1% of its value at the surface. So, we can write this as \(g' = 0.01g\). We also know that \(g' = GM/(R+h)^2\), where \(G\) is the universal gravitational constant, \(M\) is the mass of the Earth, and \(R\) is the radius of the Earth.
02

Setting up the equation

By substituting \(g' = 0.01g\) in the equation \(g' = GM/(R+h)^2\), we get \(0.01g = GM/(R+h)^2\). We also know that the value of \(g = GM/R^2\). So, let's substitute this value in our equation.
03

Solving for height \(h\)

By substituting the value of \(g\) in our equation, we get \(0.01(GM/R^2) = GM/(R+h)^2\). The \(GM\) term will cancel out from both sides, giving us \(0.01/R^2 = 1/(R+h)^2\). Solving this equation will yield \(h = 100R - R = 99R\).
04

Converting the height to radius multiples

The result \(h = 99R\) shows that the height is 99 times the radius of the Earth. However, the available options (a) 8R, (b) 9R, (c) 10R and (d) 20R are given in terms of the radius. To convert \(h = 99R\) into multiples of the radius, we can divide the height with the radius. This gives us \(h/R = 99\). Therefore, none of the options (a), (b), (c) and (d) are correct.

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