At what distance below the surface of earths acceleration due to gravity decreases by 10% of its value at the surface given radius of Earth is 6400 km?

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 equational 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

Given: gd = 90% of g i.e., `"g"_"d"/"g"` = 0.9,
R = 6400 km = 6.4 × 106 m

To find: Distance below the Earth’s surface (d)

Formula: gd = `"g"(1 - "d"/"R")`

Calculation: From formula,

`"g"_"d"/"g" = (1 - "d"/"R")`

∴ `"d"/"R" = 1 - "g"_"d"/"g"`

∴ d = R`(1 - "g"_"d"/"g")`

= 6.4 × 106 × 0.1

= 640 × 103 m

= 640 km

At distance 640 km below the surface of the Earth, value of acceleration due to gravity decreases by 10%.