Question

mass of sun is 2×10^30kg and mass of earth is 6×10^24 kg and the distance between them is 1.5×10^8 find the force between them​

Answer 1

◔ To Find :

The force between the sun and Earth.

◔ Given :

• Mass of Sun = 2 × 10³⁰ kg.

• Mass of Earth = 6 × 10²⁴ kg.

• Distance between Earth and Sun = 1.5 × 10⁸

◔ We Know :

Newton's law of Gravitation :

$\purple{\sf{\underline{\boxed{F = G\left(\dfrac{m_{1}m_{2}}{r^{2}}\right)}}}}$

Where ,

• $m_{1}$ = Mass of the Body.

• $m_{2}$ = Mass of the Body.

• $G$ = Universal Gravitational Constant.

• F = Force between the Two Bodies.

Value of Universal Gravitational Constant :

$\purple{\sf{\underline{\boxed{G = 6.67 \times 10^{-11}}}}}$

◔ Solution :

Given :

• Mass of Sun = 2 × 10³⁰ kg.

• Mass of Earth = 6 × 10²⁴ kg.

• Distance between Earth and Sun = 1.5 × 10⁸

Using the formula , and substituting the values in it ,we get :

$\purple{\sf{F = G\left(\dfrac{m_{1}m_{2}}{r^{2}}\right)}}$

$\sf{\Rightarrow F = 6.67 \times 10^{-11} \times \left(\dfrac{2 \times 10^{30} \times 6 \times 10^{24}}{(1.5 \times 10^{8})^{2}}\right)}$

$\sf{\Rightarrow F = 6.67 \times 10^{-11} \times \left(\dfrac{12 \times 10^{54}}{(2.25 \times 10^{16})}\right)}$

$\sf{\Rightarrow F = 6.67 \times \left(\dfrac{12 \times 10^{43}}{(2.25 \times 10^{16})}\right)}$

$\sf{\Rightarrow F = 6.67 \times \dfrac{12 \times 10^{27}}{2.25}}$

$\sf{\Rightarrow F = \dfrac{80.04 \times 10^{27}}{2.25}}$

$\sf{\Rightarrow F = 35.57 \times 10^{27}}$

$\purple{\sf{\therefore F = 35.57 \times 10^{27} N}}$

Hence ,the Force between Earth ans Sun is 35.57 × 10²⁷ N.

» Additional information :

• Acceleration due to gravity = $\sf{F = \dfrac{G\:M}{r^{2}}}$

• Weight = $\sf{W = na}$

• Acceleration due to gravity (Depth) = $g_{d} = g_{0}\bigg(1 - \dfrac{d}{R}\bigg)$

Answer 2

Mass of Sun = 2 × 10³⁰ kg.

Mass of Earth = 6 × 10²⁴ kg.

Distance between Earth and Sun = 1.5 × 10⁸

◔ We Know :

Newton's law of Gravitation :

\purple{\sf{\underline{\boxed{F = G\left(\dfrac{m_{1}m_{2}}{r^{2}}\right)}}}}

F=G(

r

2

m

1

m

2

)

Where ,

m_{1}m

1

= Mass of the Body.

m_{2}m

2

= Mass of the Body.

GG = Universal Gravitational Constant.

F = Force between the Two Bodies.

Value of Universal Gravitational Constant :

\purple{\sf{\underline{\boxed{G = 6.67 \times 10^{-11}}}}}

G=6.67×10

−11

◔ Solution :

Given :

Mass of Sun = 2 × 10³⁰ kg.

Mass of Earth = 6 × 10²⁴ kg.

Distance between Earth and Sun = 1.5 × 10⁸

Using the formula , and substituting the values in it ,we get :

\purple{\sf{F = G\left(\dfrac{m_{1}m_{2}}{r^{2}}\right)}}F=G(

r

2

m

1

m

2

)

$$\begin{lgathered}\\\end{lgathered}$$

$$\sf{\Rightarrow F = 6.67 \times 10^{-11} \times \left(\dfrac{2 \times 10^{30} \times 6 \times 10^{24}}{(1.5 \times 10^{8})^{2}}\right)}$$

$$\begin{lgathered}\\\end{lgathered}$$

$$\sf{\Rightarrow F = 6.67 \times 10^{-11} \times \left(\dfrac{12 \times 10^{54}}{(2.25 \times 10^{16})}\right)}$$

$$\begin{lgathered}\\\end{lgathered}$$

$$\sf{\Rightarrow F = 6.67 \times \left(\dfrac{12 \times 10^{43}}{(2.25 \times 10^{16})}\right)}$$

$$\begin{lgathered}\\\end{lgathered}$$

$$\sf{\Rightarrow F = 6.67 \times \dfrac{12 \times 10^{27}}{2.25}}$$

$$\begin{lgathered}\\\end{lgathered}$$

$$\sf{\Rightarrow F = \dfrac{80.04 \times 10^{27}}{2.25}}$$

$$\begin{lgathered}\\\end{lgathered}$$

$$\sf{\Rightarrow F = 35.57 \times 10^{27}}$$

$$\begin{lgathered}\\\end{lgathered}$$

$$\purple{\sf{\therefore F = 35.57 \times 10^{27} N}}$$

$$\begin{lgathered}\\\end{lgathered}$$

Hence ,the Force between Earth ans Sun is 35.57 × 10²⁷ N.

» Additional information :

Acceleration due to gravity = $$\sf{F = \dfrac{G\:M}{r^{2}}}$$

Weight = $$\sf{W = na}$$

Acceleration due to gravity (Depth) = $$g_{d} = g_{0}\bigg(1 - \dfrac{d}{R}\bigg)$$