Question

if the radius of the earth shrinks by $0.2\%$ without any charge in its mass, the escape velocity from the surface of the earth

• A. Increases by $0.2\%$
• B. decreases by $2.02\%$
• C. Increases by $1.01\%$
• D. decreases by $0.4\%$

Answer 1

The correct option is C Increases by $1.01\%$

Given that,

Mass of earth = M

Radius of earth = R

We know that,

The escape velocity is

$v=\sqrt{\frac{2GM}{R}}$

Now, the radius

$R^{\prime }=R-\frac{R}{0.2}$

$R^{\prime }=R-\frac{R}{20}$

$R^{\prime }=0.98R$

Now, the escape velocity is

$v^{\prime }=\sqrt{\frac{2GM}{0.98R}}$

$v^{\prime }=1.0101\sqrt{\frac{2GM}{R}}$

Now, the change in escape velocity

$=v^{\prime }-v$

$=1.0101\sqrt{\frac{2GM}{R}}-\sqrt{\frac{2GM}{R}}$

$=(1.0101-1)\sqrt{\frac{GM}{R}}$

$\frac{v^{\prime }-v}{v}\times 100=1.01\%$

Hence, the escape velocity is increases by 1.01%