Get Answers to all your Questions

header-bg qa

explain in detail

Answers (1)

There is a relationship for a body to be weightless at the equator.

That relationship is given by:

begin mathsize 12px style straight omega equals square root of straight g over straight R end root end style

where ′w′ is the angular speed,′g′ is the acceleration due to gravity and R is the radius of the Earth.

Substituting the given values in the above equation,

6400 km = 6400 x 103 m

begin mathsize 12px style straight omega equals square root of fraction numerator 10 over denominator 6400 cross times 10 cubed end fraction end root
straight omega equals square root of 1 over 640000 end root
straight omega equals 1 over 800 space rad divided by straight s end style

Posted by

Satyajeet Kumar

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE