Solveeit Logo
Login

Question

Question: What should be the velocity of earth due to rotation about its own axis so that the weight at equato...

What should be the velocity of earth due to rotation about its own axis so that the weight at equator become 3/5 of initial value. Radius of earth on equator is 6400 km

A

7.4×10−4rad/sec7.4 \times 10 ^ { - 4 } \mathrm { rad } / \mathrm { sec }

B

6.7×10−4rad/sec6.7 \times 10 ^ { - 4 } \mathrm { rad } / \mathrm { sec }

C

7.8×10−4rad/sec7.8 \times 10 ^ { - 4 } \mathrm { rad } / \mathrm { sec }

D

8.7×10−4rad/sec8.7 \times 10 ^ { - 4 } \mathrm { rad } / \mathrm { sec }

Answer

7.8×10−4rad/sec7.8 \times 10 ^ { - 4 } \mathrm { rad } / \mathrm { sec }

Explanation

Solution

Weight of the body at equator = 35\frac { 3 } { 5 } of initial weight

∴ g′=35gg ^ { \prime } = \frac { 3 } { 5 } g (because mass remains constant)

g′=g−ω2Rcos⁡2λg ^ { \prime } = g - \omega ^ { 2 } R \cos ^ { 2 } \lambda ⇒ 35g=g−ω2Rcos⁡2(0∘)\frac { 3 } { 5 } g = g - \omega ^ { 2 } R \cos ^ { 2 } \left( 0 ^ { \circ } \right)

⇒ ω2=2g5R\omega ^ { 2 } = \frac { 2 g } { 5 R } ⇒ ω=2g5R=2×105×6400×103\omega = \sqrt { \frac { 2 g } { 5 R } } = \sqrt { \frac { 2 \times 10 } { 5 \times 6400 \times 10 ^ { 3 } } }

= 7.8×10−4radsec7.8 \times 10 ^ { - 4 } \frac { \mathrm { rad } } { \mathrm { sec } }