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 cos2θ
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%.