
What is the equation that represents Gauss theorem for gravitational field?
A. $\oint {\overrightarrow g \cdot \overrightarrow {ds} } = \dfrac{m}{G}$
B. $ - \oint {\overrightarrow g \cdot \overrightarrow {ds} } = 4\pi mG$
C. $\oint {\overrightarrow g \cdot \overrightarrow {ds} } = \dfrac{m}{{4\pi G}}$
D. $ - \oint {\overrightarrow g \cdot \overrightarrow {ds} } = \dfrac{m}{G}$
Answer
232.8k+ views
Hint: According to gauss law for electric fields, the net electric flux through any closed surface is equal to $\dfrac{1}{{{\varepsilon _0}}}$ times the total electric charge $q$ enclosed, by the surface.
That is $\oint {E.ds} = \dfrac{q}{{{\varepsilon _0}}}$
Where the electric field, $E$ is given by the equation
$E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{{{r^2}}}$
Gravitational field is given by the equation
$g = G\dfrac{m}{{{r^2}}}$
By comparing the equation for electric field and gravitational field we can arrive at the gauss law for gravitational field.
Complete step by step answer:
According to gauss law for electric fields, the net electric flux through any closed surface is equal to $\dfrac{1}{{{\varepsilon _0}}}$ times the total electric charge $q$ enclosed, by the surface.
That is $\oint {E.ds} = \dfrac{q}{{{\varepsilon _0}}}$ (1)
Where the electric field is given by the equation
$E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{{{r^2}}}$ (2)
Similarly, we can write the gauss law for gravitational field.
We know gravitational field is given by the equation
$g = G\dfrac{m}{{{r^2}}}$ (3)
Now compare equation (3) and (2)
We can see that $q$in electric field is analogous to$m$ in gravitational field.
Also, the constants $\dfrac{1}{{4\pi {\varepsilon _0}}}$ in electric field is analogous to the gravitational constant $G$in gravitational field. Now we can equate these constants.
$
\dfrac{1}{{4\pi {\varepsilon _0}}} = G \\
\dfrac{1}{{{\varepsilon _0}}} = 4\pi G \\
$
So, let us replace the gravitational analogues in the equation (1)
$
\oint {E.ds} = \dfrac{1}{{{\varepsilon _0}}} \times q \\
\oint {g.ds} = 4\pi G \times m \\
$
The gravitational force is always an attractive force. Hence, we have to consider a negative sign in the equation of gauss law for the gravitational field.
Thus, our final answer is option B, $ - \oint {\overrightarrow g \cdot \overrightarrow {ds} } = 4\pi mG$
Note: It is important to note that Gravitational force is always attractive in nature,but $4\pi mG$ is always a positive quantity and hence we include a negative sign in the left hand side of the equation to maintain the sign convention(negative field for attraction).
That is $\oint {E.ds} = \dfrac{q}{{{\varepsilon _0}}}$
Where the electric field, $E$ is given by the equation
$E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{{{r^2}}}$
Gravitational field is given by the equation
$g = G\dfrac{m}{{{r^2}}}$
By comparing the equation for electric field and gravitational field we can arrive at the gauss law for gravitational field.
Complete step by step answer:
According to gauss law for electric fields, the net electric flux through any closed surface is equal to $\dfrac{1}{{{\varepsilon _0}}}$ times the total electric charge $q$ enclosed, by the surface.
That is $\oint {E.ds} = \dfrac{q}{{{\varepsilon _0}}}$ (1)
Where the electric field is given by the equation
$E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{{{r^2}}}$ (2)
Similarly, we can write the gauss law for gravitational field.
We know gravitational field is given by the equation
$g = G\dfrac{m}{{{r^2}}}$ (3)
Now compare equation (3) and (2)
We can see that $q$in electric field is analogous to$m$ in gravitational field.
Also, the constants $\dfrac{1}{{4\pi {\varepsilon _0}}}$ in electric field is analogous to the gravitational constant $G$in gravitational field. Now we can equate these constants.
$
\dfrac{1}{{4\pi {\varepsilon _0}}} = G \\
\dfrac{1}{{{\varepsilon _0}}} = 4\pi G \\
$
So, let us replace the gravitational analogues in the equation (1)
$
\oint {E.ds} = \dfrac{1}{{{\varepsilon _0}}} \times q \\
\oint {g.ds} = 4\pi G \times m \\
$
The gravitational force is always an attractive force. Hence, we have to consider a negative sign in the equation of gauss law for the gravitational field.
Thus, our final answer is option B, $ - \oint {\overrightarrow g \cdot \overrightarrow {ds} } = 4\pi mG$
Note: It is important to note that Gravitational force is always attractive in nature,but $4\pi mG$ is always a positive quantity and hence we include a negative sign in the left hand side of the equation to maintain the sign convention(negative field for attraction).
Recently Updated Pages
Circuit Switching vs Packet Switching: Key Differences Explained

Dimensions of Pressure in Physics: Formula, Derivation & SI Unit

JEE General Topics in Chemistry Important Concepts and Tips

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

Waves Class 11 Physics Chapter 14 CBSE Notes - 2025-26

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

