If both the mass and the radius of the earth | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The acceleration due to gravity $g$ is given by the formula $g = \frac{GM}{R^2}$,where $G$ is the gravitational constant,$M$ is the mass of the Ea

Vedclass · Accepted Answer

Increase by $1\%$

Vedclass · Answer

(B) The acceleration due to gravity $g$ is given by the formula $g = \frac{GM}{R^2}$,where $G$ is the gravitational constant,$M$ is the mass of the Earth,and $R$ is the radius of the Earth.Taking the natural logarithm on both sides: $\ln(g) = \ln(G) + \ln(M) - 2\ln(R)$.Differentiating both sides to find the fractional change: $\frac{dg}{g} = \frac{dM}{M} - 2\frac{dR}{R}$.Given that the mass $M$ decreases by $1\%$,so $\frac{dM}{M} = -0.01$.Given that the radius $R$ decreases by $1\%$,so $\frac{dR}{R} = -0.01$.Substituting these values into the equation: $\frac{dg}{g} = (-0.01) - 2(-0.01) = -0.01 + 0.02 = +0.01$.This represents an increase of $1\%$ in the value of $g$.

If both the mass and the radius of the earth decrease by $1\%$,the value of the acceleration due to gravity will

Similar Questions

The radii of two planets $A$ and $B$ are $R$ and $4R$ and their densities are $\rho$ and $\rho/3$ respectively. The ratio of acceleration due to gravity at their surfaces $(g_A : g_B)$ will be

The moon's radius is $1/4$ that of the earth and its mass is $1/80$ times that of the earth. If $g$ represents the acceleration due to gravity on the surface of the earth,then the acceleration due to gravity on the surface of the moon is:

If the density of the earth is doubled keeping its radius constant,then the acceleration due to gravity will be........ $m/s^2$. $(g = 9.8\,m/s^2)$

The height $h$ above the earth's surface at which the value of acceleration due to gravity $g$ becomes $\frac{g}{3}$ is ($R=$ radius of the earth).

Write the difference between $G$ and $g$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module