At what height above the earths surface is the acceleration due to gravity is 1% less than its value at the surface?

India's Super Teachers for all govt. exams Under One Roof

Enroll For Free Now

CONCEPT:

  • Gravitational acceleration: It is the acceleration gained by an object due to gravitational force. 
    • ​Its SI unit is m/s2.
    • It has both magnitude and direction, hence, it’s a vector quantity.
    • Acceleration due to gravity is represented by g.
    • The standard value of g on the surface of the earth at sea level is 9.8 m/s2.

Acceleration due to gravity on the surface is given:

g1 = GM/R2

Acceleration due to gravity above the surface is given:

g2 = GM/(R + H)2

Where, R = Radius of the earth (6400 km), and H = height in meters

CALCULATION:

From the above discussion it's clear that

\(\frac{{{g_2}}}{{{g_1}}} = \frac{{GM/{{(R + H)}^2}}}{{GM/{R^2}}}\)

\(\frac{{{g_2}}}{{{g_1}}} = \frac{{{R^2}}}{{{{(R + H)}^2}}}\)

\(\frac{{{g_2}}}{{9.8}} = \frac{{{{(6.4 \times {{10}^6})}^2}}}{{{{(6.4 \times {{10}^6} + 0.01 \times {{10}^6})}^2}}}\)

g2 = 9.77 m/s2

The correct option is 9.77 m/s2

Ace your Gravitation preparations for The gravitational constant with us and master Physics for your exams. Learn today!

India’s #1 Learning Platform

Start Complete Exam Preparation

Daily Live MasterClasses

Practice Question Bank

Mock Tests & Quizzes

Get Started for Free Download App

Trusted by 3.4 Crore+ Students

Neuester Beitrag

Stichworte