Question

15. The radius of
radius of the earth is 6400 km and
10m/s. In order that a body of 5 kg
lehs zero at the equator, the angular
speed of the earth is
a 1/80 radian/sec (b)1/400 radian/sec
al 1/800 radian/sec(d) 1/1600 radian/sec​

Answer 1

$\huge{\bold{\underline{\underline{\tt Correct \: question:-}}}}$

The radius and acceleration due to gravity of the earth is 6400 km and 10m/s². In order that a body of 5 kg weighs zero at the equator, the angular speed of the earth is

(a) 1/80 radian/sec

(b)1/400 radian/sec

(c) 1/800 radian/sec

(d) 1/1600 radian/sec

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Radius of earth (R) = 6400Km.
• Mass of the body (m) = 5kg.

$\huge{\bold{\underline{Explanation:-}}}$

$\rule{300}{1.5}$

Derivation:-

If a body is at height "h". The Gravitational Force will be provided by the Centripetal Force. (Acting on the body)

Now,

$\large{\boxed{\tt F_{(Gravitational \: Force)} = F_{(Centripetal \: Force)}}}$

Applying The formulas,

$\large{\tt \leadsto \dfrac{GMm}{(R + h)^2} = m \omega^2 r}$

(Here the Distance is taken as (R + h) because it is assumed to that body is at height "h")

Now,

$\large{\tt \leadsto \dfrac{GM\cancel{m}}{(R + h)^2} = \cancel{m} \omega^2 r}$

$\large{\tt \leadsto \dfrac{GM}{(R + h)^2} = \omega^2 r}$

$\large{\tt \leadsto \omega^2 = \dfrac{GM}{(R + h)^2 \times (R + h)}}$

$\large{\tt \leadsto \omega^2 = \dfrac{GM}{(R + h)^3}}$

$\large{\boxed{\tt \omega = \sqrt{\dfrac{GM}{(R + h)^3}}}}$

$\rule{300}{1.5}$

$\rule{300}{1.5}$

As in the question specified The object is at earth's surface,

Therefore

$\large{ \tt{ \: h \: \approx 0 \: \: \: \rightarrow (h < < R)}}$

Now, The Former becomes,

$\large{\tt \leadsto \omega^2 = \dfrac{GM}{(R)^3}}$

But from "G"(Universal Gravitational constant) and "g" (Acceleration due to gravity) relation,

$\large{\tt \leadsto GM = gR^2}$

Substituting in the Angular velocity Formula,

$\large{\tt \leadsto \omega^2 = \dfrac{gR^2}{(R)^3}}$

$\large{\tt \leadsto \omega^2 = \dfrac{g\cancel{R^2}}{\cancel{(R)^3}}}$

$\large{\underline{\underline {\tt \leadsto \omega = \sqrt{\dfrac{g}{R}}}}}$

Substituting the values,

$\large{\tt \leadsto \omega = \sqrt{\dfrac{10}{6400 \times 1000}}}$

As the Radius is given in Km, it needs to be converted into Meters

$\large{\tt \leadsto \omega = \sqrt{\dfrac{\cancel{10}}{6400 \times \cancel{1000}}}}$

$\large{\tt \leadsto \omega = \sqrt{\dfrac{1}{6400 \times 100}}}$

Now,

$\large{\tt \leadsto \omega = \dfrac{1}{80 \times 10}}$

$\huge{\boxed{\boxed{\tt \omega = \dfrac{1}{800}\: rad/sec}}}$

So, the Angular velocity of the earth should be 1/800 rad/sec.

$\rule{300}{1.5}$