Question

If diameter is reduced to half from present diameter of earth, it's mass becomes 1/8 . what is the value of g on the earth at this situation? ​

Answer 1

since the diameter is half and mass is 1/8 of the original on the the equation would be

$d = \frac{r}{2}$

$\frac{d}{2} = \frac{r}{4}$

and mass is $\frac{1}{8}$

the f (force due to gravity would be)

$f = G× \frac{16m}{8r^2}$

since the Earth exert equal forces on each thing so there is no need of mass of other object.

$f = G× \frac{m}{r^2}$

since the value if there is no change would be this so we can write the equation like this

$f = 2× f_1$

there $f_1$is the value above

hence the original force now would be. the half of present gravitational acceleration.

so g= 4.9

Answer 2

Let the original diameter of the Earth be d₁ .

Let the changed diameter of the Earth be d₂ .

d₁ = 2 d₂ because the diameter is halved .

⇒ 2 r₁ = 4 r₂

⇒ r₁ = 2 r₂

Hence the original diameter is twice the changed .

r₁ = 2 r₂

⇒ r₂ = r₁/2

Let the original mass be m₁ .

The changed mass is m₁/8 and the mass becomes 1/8 of the original .

m₂ = m₁/8

Let the mass of the Earth be m .

$\mathsf{Original\:g=\dfrac{G(m_1)m}{r_1^2}}$

$\mathsf{Changed\:g=\dfrac{G(m_2)(m)}{r_2^2}}\\\\\implies \mathsf{\dfrac{G(\dfrac{1}{8m_1})(m)}{(2r_1)^2}}\\\\\implies \mathsf{\dfrac{G(m_1)(m)}{r_1^2}\times \dfrac{2^2}{8}}\\\\\implies \mathsf{\dfrac{G(m_1)(m)}{r_1^2}\times \dfrac{1}{2}}$

Hence the value of g becomes half of the original value .