The universal gravitational constant is measured experimentally using Cavendish’s experiment. This experiment is a very precise experiment which gives the value of G as $(6.6742 \pm 0.0010) \times 10^{- 11} N . m^{2} / k g^{2} .$ Since the universal gravitational constant has some error in it, using it if we find escape speed and acceleration due to gravity on Earth’s surface, our values will have error in it.(Assume mass of Earth $= 6.0000 \times 10^{24}$ kg & radius of Earth= $6.4000 \times 10^{6} m$ ). The measurement of mass of Earth has an error of $0.01 %$ and radius of Earth has an error of $0.02 %$
Choose the CORRECT option(s)

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

Maximum percentage error in acceleration due to gravity $g = 0.025$

Maximum percentage error in acceleration due to gravity $g = 0.065$

Value of acceleration due to gravity (approx. in $m / s^{2}$ ) $= 9.78$

Value of acceleration due to gravity (approx.in $m / s^{2}$ ) $= 8.78$

answer is A, C.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$g = \frac{G M}{R^{2}} = \frac{6.6742 \times 10^{- 11} \times 6 \times 10^{24}}{{6.4}^{2} \times 10^{12}} = 9.77666 \frac{d g}{g} = \frac{d G}{G} + \frac{d M}{M} + \frac{2 d R}{R} = \frac{0.001}{6.6742} + 0.0001 + 2 \times 0.0002 = 0.00065 ∴ g = (9.77767 \pm 0.0007) m / s^{2}$