Question

If a planet existed whose mass and radius were both half of that of Earth, what would be the acceleration due to gravity at the surface of the planet in terms of that on the surface of Earth ?

Answer 1

Explanation:

If the

mass of planet = 1/2of earth's mass

radius. = 1/2of earth's radius

gravity. = m/ r^2

=

Less the radius and mass less it will have gravity.

Answer 2

Answer:

The acceleration due to gravity on the planet would be twice that of the earth.

Explanation:

We know that,

$\bullet\sf\:The\:mass\:of\:earth\:=\:M\\\\\\\bullet\sf\:Radius\:of\:the\:earth\:=\:R\\\\\\\bullet\sf\:g\:=\:Acceleration\:due\:to\:gravity$

Let the mass and radius of the planet be,

$\bullet\sf\:M_p\:=\:Mass\:of\:the\:planet\\\\\\\bullet\sf\:R_p\:=\:Radius\:of\:the\:planet\\\\\\\bullet\sf\:g_p\:=\:Acceleration\:due\:to\:gravity\:on\:planet$

We have given that,

$\longrightarrow\sf\:M_p\:=\:\dfrac{M}{2}\\\\\\\longrightarrow\sf\:R_p\:=\:\dfrac{R}{2}$

We know that,

$\boxed{\pink{\sf\:g\:=\:\dfrac{G\:M}{R^2}}}\\\\\\\therefore\sf\:g_p\:=\:\dfrac{G\:M_p}{R^2_p}\:\:\:-\:-\:[\:G\:is\:constant\:]\\\\\\\implies\sf\:g_p\:=\:\dfrac{G\:\dfrac{M}{2}}{\dfrac{R^2}{4}}\\\\\\\implies\sf\:g_p\:=\:\dfrac{G\:M}{\cancel2}\:\times\:\dfrac{\cancel4}{R^2}\\\\\\\implies\sf\:g_p\:=\:2\:\times\:G\:\times\:\dfrac{M}{R^2}\\\\\\\implies\boxed{\sf\:g_p\:=\:2\:\times\:\dfrac{G\:M}{R^2}}}$

The acceleration due to gravity on the planet would be twice that of the earth.