Question

suppose a planet exist whose mass and radius both are half of that of earth.calculate the acceleration due to gravity on the surface of this planet.

Answer 1

"Therefore the acceleration due to gravity on the planet is [tex]19.6\frac { m }{ { s }^{ 2 } }[/tex]

Solution:

We know that the "acceleration due to gravity" formula is

$g=\frac { GM }{ R^{ 2 } }$

Where, M is mass of planet and R is its radius

In earth the "acceleration due to gravity" value $g^{\prime}=9.8 \mathrm{m} / \mathrm{s}^{2}$

On planet g' is

${ g }^{ \prime}=\frac { GM }{ { R }^{ 2 } } =\frac { GM }{ 2{ \left( \frac { R }{ 2 } \right)}^{ 2 } } =\frac { 2GM }{ { R }^{ 2 } } =2{ g }$

Thus, "acceleration due to gravity" on the planet is,

$g^{\prime}=2 g=2 \times 9.8=19.6 \mathrm{m} / \mathrm{s}^{2}$"

Answer 2

know that the "acceleration due to gravity" formula is

g=\frac { GM }{ R^{ 2 } }g=R2GM

Where, M is mass of planet and R is its radius

In earth the "acceleration due to gravity" value g^{\prime}=9.8 \mathrm{m} / \mathrm{s}^{2}g′=9.8m/s2

On planet g' is

{ g }^{ \prime}=\frac { GM }{ { R }^{ 2 } } =\frac { GM }{ 2{ \left( \frac { R }{ 2 } \right)}^{ 2 } } =\frac { 2GM }{ { R }^{ 2 } } =2{ g }g′=R2GM=2(2R)2GM=R22GM=2g

Thus, "acceleration due to gravity" on the planet is,

g^{\prime}=2 g=2 \times 9.8=19.6 \mathrm{m} / \mathrm{s}^{2}g′=2g=2×9.8=19.6m/s2 "