Question

At what height (in km) over the Earth’s pole the free-fall acceleration decreases by one percent?

Answer 1

Hint: The problem is solved using Newton’s law of gravitation, which is based on Kepler’s law of planetary motion. During free fall, the body is subjected to only gravity. $g=\dfrac{GM}{R^2}$ gravity at any height is given by $g_{h}= g(1+\dfrac{h}{R})^{-2}$ when $h << R$
$\Rightarrow g_{h}=\dfrac{g}{1-\dfrac{2h}{R}}$

Formula used: $g=\dfrac{GM}{R^2}$ gravity at any height is given by $g_{h}= g(1+\dfrac{h}{R})^{-2}$ when $ h << R $
$\Rightarrow g_{h}=\dfrac{g}{1-\dfrac{2h}{R}}$

Complete step-by-step solution -
From gravitational law, we know $g=\dfrac{GM}{R^2}$ we also know that gravity at any height is given by $g_{h}= g(1+\dfrac{h}{R})^{-2}$ where$g_{h}$ the gravity at $h$ and $R$ is the radius of the earth.
When $h << R$, using binomial expansion
$g_{h}=g\times(1-\dfrac{2h}{R})$
since \[g\] decreases by 1%
$g_{h}=\dfrac{99}{100} \times g$
$\dfrac{99}{100}\times g=g\times(1-\dfrac{2h}{R})$
$1-\dfrac{99}{100}=\dfrac{2h}{R}$
$h=\dfrac{R}{200}$
We know that radius of earth is \[6400km\]
Then,$h=\dfrac{6400}{200}=32km$
Therefore at a distance 32km above the poles the free fall of acceleration decreases by 1%

Additional Information:
The above law was given by Newton and is also called as Newton’s law of gravitational law, which states that every particle attracts every other particle in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. This is a general physical law derived from empirical observations by what Isaac Newton called inductive reasoning. This was the basis Kepler's laws of planet motion are three scientific laws describing the motion of planets around the Sun. The law states that:
The orbit of a planet is an ellipse with the sun at one of the two foci.
A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.
The ratio of the square of an object’s orbital period with the cube of the semi-major of its orbit is the same for all objects orbiting the same primary.

Note: Remember the formula used. The problem is solved using Newton’s law of gravitation, which is based on Kepler’s law of planetary motion. Remember that the radius of earth is \[6400km\], which is a universal constant.