At what height above the earths surface would the value of acceleration due to gravity be one-fourth of what is on the surface radius of the earth is 6400km?

1. Newton's law of gravitation:

(i) Gravitational force acting between to bodies, F∝m1m2r2 or F=Gm1m2r2 where G=6.67×1011N m2kg2 is the universal gravitational constant.

(ii) In vector form: F→12=Gm1m2r2r^12 & F→21=Gm1m2r2r^21

2. Gravitational field:

(i) Gravitational field is related to the force as, E→=F→m

(ii) The field produced by a point mass is given by, E→=GMr2r^

3. Variation of Acceleration due to Gravity:

(i) Acceleration due to gravity at height h from the surface gh=GMeRe+h2≃g1-2hRe

(ii) Acceleration due to gravity at depth d from the surface, gd=g1-dRe

(iii) The eqitorial radius is about 21 km longer than its polar radius. Hence gpole>gequator

(iv) Acceleration due to gravity at latitude θ, g'=g-Rω2 cos2⁡θ

4. Escape velocity:

It is the speed required from the surface of a planet to get out of the influence of the planet. For earth, ve=2GMeRe=2gRe

5. Satellite in a circular orbit:

(i) Satellite orbital Velocity, v0=GMeRe+h12

(ii) Time period of satellite, T=2πRe+h32GMe

(iii) Potential energy of a Satellite: P.E.=-GMem(Re+h), Kinetic energy: K.E.=GMem2(Re+h) and total energy E=-GMem2(Re+h)

6. Kepler’s Laws:

(i) All planets move in elliptical orbits with the Sun at one of the focal points

(ii) The line joining the sun and a planet sweeps out equal areas in equal intervals of time.

(iii) The square of the orbital period of a planet is proportional to the cube of the semi-major axis of the elliptical orbit of the planet. The time period T and radius R of the circular orbit of a planet about the sun are related as T2=4π2GMsR3

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Answer

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Hint: In order to answer this question, we will let the new gravity be $g`$ in both cases i.e.. half of $g$and one-fourth of $g$ respectively. And then we will apply the formula to find the height above the earth’s surface in the terms of radius of the earth and the gravity to find the height in the given both cases.

Complete step by step answer:

Given that- Radius of the earth is, $R = 6400km$(i) Let g is the acceleration due to gravity.Let $g`$ be the half of $g$ .So, $g` = \dfrac{g}{2}$Now, for height, $h$ above the half of the earth’s surface, the formula is:$\because g` = g{(\dfrac{R}{{R + h}})^2}$where, $R$ is the radius of the earth.Now, we will put $\dfrac{g}{2}$ instead of $g`$ , as $g` = \dfrac{g}{2}$ .$ \Rightarrow \dfrac{g}{2} = g{(\dfrac{R}{{R + h}})^2}$Cancel out $g$ from both the sides:$\Rightarrow \dfrac{R}{{R + h}} = \dfrac{1}{{\sqrt 2 }} \\\Rightarrow \sqrt 2 R = R + h \\\Rightarrow h = \sqrt 2 R - R \\ $Now, we will take $R$ as common from the above equation and as we know, $\sqrt 2 = 1.414$ :$\Rightarrow h = (\sqrt 2 - 1)R \\\Rightarrow h = (1.414 - 1)R \\\Rightarrow h = 0.414 \times 6400 \\\Rightarrow h\, = 2649.6\,km \\\therefore h \approx 2650\,km \\ $

Hence, the height above the half of its value of the earth’s surface would cause the acceleration due to gravity to be $2650\,km$.

(ii) Again, $g`$ be the one- fourth of $g$ .So, $g` = \dfrac{g}{4}$For height, $h$ above the one-fourth of the earth’s surface:$\because g` = g{(\dfrac{R}{{R + h}})^2}$where, R is the radius of the earth.Now, we will put $\dfrac{g}{4}$ instead of $g`$ , as $g` = \dfrac{g}{4}$ .$\Rightarrow \dfrac{g}{4} = g{(\dfrac{R}{{R + h}})^2} \\\Rightarrow \dfrac{R}{{R + h}} = \dfrac{1}{{\sqrt 4 }} \\\Rightarrow R + h = 2R \\\Rightarrow h = 2R - R \\\Rightarrow h = R $We will put the value of $R$ that is given.$\therefore h = 6400\,km$

Hence, the height above the one-fourth of its value of the earth’s surface would cause the acceleration due to gravity to be $6400\,km$.

Note: The distance between the Earth's centre and a location on or near its surface is measured in Earth radius. The radius of an Earth spheroid, which approximates the figure of Earth, ranges from nearly \[6,378km\] to nearly \[6,357km\] . In astronomy and geophysics, a nominal Earth radius is sometimes used as a unit of measurement, with the equatorial value suggested by the International Astronomical Union.