The mass of the sun is 1.99×1030kg and its distance to the Earth is 1.50×1011m. What is the gravitational force of the sun on the earth?

Answers

Answer 1

Answer: In order to calculate the gravitational force of the two bodies we use the formula which is expressed as:

F = GMm/R²

where G = 6.67 x 10^-11 in SI unit, M and m are the mass of the two bodies and R is the distance between them.

F = 6.67 x 10^-11 (1.99×10^30) (6×10^24) / (1.50×10^11)²
F = 3.53×10^22N

Answer 2

Answer:

The gravitational force of attraction acting on the earth due to sun is $\boxed{3.539*{{10}^(22)}\,{\text{N}}}$ or $\boxed{3.54*{{10}^(22)}\,{\text{N}}}$ .

Further Explanation:

The Newton’s law of Gravitation states that the gravitational force of attraction acting between two bodies is directly proportional to the product of the mass of the two bodies and inversely proportional to the square of the distance between the two bodies.

Given:

The mass of the sun is $1.99*{10^(30)}\,{\text{kg}}$ .

The distance between the Earth and the sun is $1.50*{10^(11)}\,{\text{m}}$ .

Concept:

The gravitational force of attraction between the two bodies can be represented mathematically as:

$F=\frac{{G{M_1}{M_2}}}{{{r^2}}}$

Here, $F$ is the gravitational force of attraction, $G$ is the gravitational constant, ${M_1}$ is the mass of the first body, ${M_2}$ is the mass of the second body and $r$ is the distance between them.

Since the two bodies are Sun and the Earth so the masses of the two are considered.

The mass of the Earth is $6* {10^(24)}\,{\text{kg}}$ and the value of Gravitational constant is $6.67*{10^( - 11)}\,{{{\text{N}}\cdot{{\text{m}}^{\text{2}}}}\mathord{\left/{\vphantom{{{\text{N}}\cdot{{\text{m}}^{\text{2}}}}{{\text{k}}{{\text{g}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace} {{\text{k}}{{\text{g}}^{\text{2}}}}}$ .

Substitute $6.67*{10^( - 11)}\,{{{\text{N}}\cdot{{\text{m}}^{\text{2}}}}\mathord{\left/{\vphantom{{{\text{N}}\cdot{{\text{m}}^{\text{2}}}}{{\text{k}}{{\text{g}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace} {{\text{k}}{{\text{g}}^{\text{2}}}}}$ for $G$ , $1.99*{10^(30)}\,{\text{kg}}$ for ${M_1}$ , $6* {10^(24)}\,{\text{kg}}$ for ${M_2}$ and $1.50*{10^(11)}\,{\text{m}}$ for $r$ in above equation.

$\begin{aligned}F&=\frac{{\left({6.67*{{10}^(-11)}\,}\right)*\left({1.99*{{10}^(30)}\,{\text{kg}}}\right)*\left({6*{{10}^(24)}\,{\text{kg}}}\right)}}{{{{\left({1.50*{{10}^(11)}\,{\text{m}}}\right)}^2}}}\n&=\frac{{7.964*{{10}^(44)}}}{{2.25*{{10}^(22)}}}\n&=3.539*{10^(22)}\,{\text{N}}\n\end{aligned}$

Therefore, the gravitational force of attraction acting on the earth due to sun is $\boxed{3.539*{{10}^(22)}\,{\text{N}}}$ or $\boxed{3.54* {{10}^(22)}\,{\text{N}}}$ .

Learn More:

1. Calculate the total force on the earth due to Venus, Jupiter and Saturn brainly.com/question/2887352

2. A rocket being thrust upward as the force of the fuel being burned pushes downward brainly.com/question/11411375

3. A 50-kg meteorite moving at 1000 m/s strikes earth. Assume the velocity is along the line brainly.com/question/6536722

Answer details:

Grade: High school

Subject: Physics

Chapter: Newton’s law of Gravitation

Keywords:

Gravitational force, attraction, newton’s law, earth, sun, mass of the sun, distance, 1.99*10^30 kg, 1.50*10^11 m, 6.67x10^-11, gravitational constant.

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