Question

Given that a planets mass and diameter are 3 and 6 times thaf of earth respectively calculate the ratio between the acceleration due to gravity on that planet and that on earth​

Answer 1

$\red{\large\underline{\underline\mathtt{Question:}}}$

Given that a Planet's mass and diameter are 3 and 6 times thaf of earth respectively. Calculate the ratio between the acceleration due to gravity on that planet and that on earth.

$\purple{\large\underline{\underline\mathtt{To\:find:}}}$

The ratio between the. acceleration due to gravity on that planet and that on earth.

$\blue{\large\underline{\underline\mathtt{We\:know:}}}$

Radius of Earth = $R_{1} \rightarrow \mathtt{6.4 \times 10^{6}m}$

Mass of Earth = $M_{1} \rightarrow \mathtt{6 \times 10^{24}kg}$

$\mathtt{ Radius = \dfrac{Diameter}{2}}$

$\mathtt{g = \dfrac{GM}{R^{2}}}$

where,

g = Acceleration due to Gravity

G = Universal Gravitational constant

R = Distance

$G = 6.67 \times 10^{-11}$

$\green{\large\underline{\underline\mathtt{Given:}}}$

Diameter of the Planet = $D_{2}$ = $3 \times of Earth$

$\Rightarrow D_{2} = 6 \times (2 \times 6.4 × 10^{6})$

$\therefore D_{2} = 76.8 ×\times 10^{6}$

We know ,

$\Rightarrow Radius = \dfrac{76.8 \times 10^{6}}{2}$

$\Rightarrow Radius = \dfrac{\cancel{76.8} \times 10^{6}}{\cancel{2}}$

$\Rightarrow R_{2} \rightarrow Radius = 38.4 \times 10^{6}$

Mass of the Planet = $M_{2} = 3 \times of Earth$

$\Rightarrow M_{2} = 3 \times 6 \times 10^{24}$

$\Rightarrow M_{2} = 18 \times 10^{24}$

$\red{\large\underline{\underline\mathtt{Solution:}}}$

Acceleration due to Gravity on thr surface of Earth

We know,

$\mathtt{g = \dfrac{GM}{R^{2}}}$

Putting the value in the formula we , get:-

$\Rightarrow g = \dfrac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{(6.4 \times 10^{6})^{2}}$

$\Rightarrow g = \dfrac{6.67 \times 6 \times 10^{13}}{6.4 \times 6.4\times 10^{12}}$

$\Rightarrow g = \dfrac{6.67 \times 6 \times \cancel{10^{13}}}{6.4 \times 6.4\times \cancel{10^{12}}}$

$\Rightarrow g = \dfrac{6.67 \times 6 \times 10}{6.4 \times 6.4}$

$\Rightarrow g = 9.8ms^{-2} (Approx.)$

• Acceleration due to gravity in the planet x.

We know,

$\mathtt{g = \dfrac{GM}{R^{2}}}$

Putting the value in the Formula, we get:-

$\Rightarrow g = \dfrac{6.67 \times 10^{-11} \times 18 \times 10^{24}}{(38.4 \times 10^{6})^{2}}$

$\Rightarrow g = \dfrac{6.67 \times 18 \times 10^{13}}{(38.4 \times 10^{6})^{2}}$

$\Rightarrow g = \dfrac{6.67 \times 18 \times 10^{13}}{38.4 \times 38.4\times 10^{12}}$

$\Rightarrow g = \dfrac{6.67 \times 18 \times \cancel{10^{13}}}{38.4 \times 38.4\times \cancel{10^{12}}}$

$\Rightarrow g = \dfrac{6.67 \times 18 \times 10}{38.4 \times 38.4}$

$\Rightarrow g = 0.81$

Ratio of the Acceleration due to gravity on planet X and Earth.

$\mathtt{g = \dfrac{GM_{x}}{R^{2}} : g = \dfrac{GM_{e}}{R_{e}^{2}}}$

$\Rightarrow 0.81 : 9.8$

$\Rightarrow 1 : 12$

${\boxed{\therefore Ratio = 1 : 12}}$

Hence , the the ratio of Acceleration due to Gravity on the planet x and Planet earth is 1 : 12....

______________________________________

Answer 2

$\huge\bold\red\bigstar{Question}$

Given that a Planet's mass and diameter are 3 and 6 times thaf of earth respectively. Calculate the ratio between the acceleration due to gravity on that planet and that on earth.

$\purple{\large\underline{\underline\mathtt{To\:find:}}}$

The ratio between the. acceleration due to gravity on that planet and that on earth.

$\blue{\large\underline{\underline\mathtt{We\:know:}}}$

Radius of Earth = R_{1} $\rightarrow \mathtt{6.4 \times 10^{6}m}$R

1

→6.4×10

6

m

Mass of Earth = M_{1}$\rightarrow \mathtt{6 \times 10^{24}kg}$M

1

→6×10

24

kg

$\mathtt{ Radius = \dfrac{Diameter}{2}}$

$\mathtt{g = \dfrac{GM}{R^{2}}}$

where,

g = Acceleration due to Gravity

G = Universal Gravitational constant

R = Distance

G = 6.67 \times 10^{-11}G=6.67×10

−11

$\green{\large\underline{\underline\mathtt{Given:}}}$

Diameter of the Planet = D_{2}D

2

= 3 \times of Earth3×ofEarth

$\Rightarrow D_{2} = 6 \times (2 \times 6.4 × 10^{6})$

$\therefore D_{2} = 76.8 ×\times 10^{6}$

We know ,

$\Rightarrow Radius = \dfrac{76.8 \times 10^{6}}{2}$

$\Rightarrow Radius = \dfrac{\cancel{76.8} \times 10^{6}}{\cancel{2}}$

$\Rightarrow R_{2} \rightarrow Radius = 38.4 \times 10^{6}$

Mass of the Planet = M_{2} = 3 \times of EarthM

2

=3×ofEarth

$\Rightarrow M_{2} = 3 \times 6 \times 10^{24}$⇒M

2

=3×6×10

24

$\Rightarrow M_{2} = 18 \times 10^{24}$⇒M

2

=18×10

24

$\red{\large\underline{\underline\mathtt{Solution:}}}$

Acceleration due to Gravity on thr surface of Earth

We know,

$\mathtt{g = \dfrac{GM}{R^{2}}}$

Putting the value in the formula we , get:-

$\Rightarrow g = \dfrac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{(6.4 \times 10^{6})^{2}}$⇒g=

(6.4×10

6

)

2

6.67×10

−11

×6×10

24

$\Rightarrow g = \dfrac{6.67 \times 6 \times 10^{13}}{6.4 \times 6.4\times 10^{12}}$⇒g=

6.4×6.4×10

12

6.67×6×10

13

$\Rightarrow g = \dfrac{6.67 \times 6 \times \cancel{10^{13}}}{6.4 \times 6.4\times \cancel{10^{12}}}$⇒g=

6.4×6.4×

10

12

6.67×6×

10

13

$\Rightarrow g = \dfrac{6.67 \times 6 \times 10}{6.4 \times 6.4}$⇒g=

6.4×6.4

6.67×6×10

$\Rightarrow$g = 9.8ms^{-2} (Approx.)⇒g=9.8ms

−2

(Approx.)

Acceleration due to gravity in the planet x.

We know,

$\mathtt{g = \dfrac{GM}{R^{2}}}$g=

R

2

GM

Putting the value in the Formula, we get:-

$\Rightarrow g = \dfrac{6.67 \times 10^{-11} \times 18 \times 10^{24}}{(38.4 \times 10^{6})^{2}}$⇒g=

(38.4×10

6

)

2

6.67×10

−11

×18×10

24

$\Rightarrow g = \dfrac{6.67 \times 18 \times 10^{13}}{(38.4 \times 10^{6})^{2}}$

$\Rightarrow g = \dfrac{6.67 \times 18 \times 10^{13}}{38.4 \times 38.4\times 10^{12}}$

$\Rightarrow g = \dfrac{6.67 \times 18 \times \cancel{10^{13}}}{38.4 \times 38.4\times \cancel{10^{12}}}$

$\Rightarrow g = \dfrac{6.67 \times 18 \times 10}{38.4 \times 38.4}$

$\Rightarrow$ g = 0.81⇒g=0.81

Ratio of the Acceleration due to gravity on planet X and Earth.

$\mathtt{g = \dfrac{GM_{x}}{R^{2}} : g = \dfrac{GM_{e}}{R_{e}^{2}}}$

$\Rightarrow$ 0.81 : 9.8⇒0.81:9.8

$\Rightarrow$ 1 : 12⇒1:12

${\boxed{\therefore Ratio = 1 : 12}}$

Hence , the the ratio of Acceleration due to Gravity on the planet x and Planet earth is 1 : 12....

______________________________________