
If acceleration due to gravity on the surface of the earth is \[9.8m{s^ {- 2}}\], what will be the acceleration due to gravity on the surface of a plant whose mass and radius both are two times the acceleration corresponding quantities for the earth?
A) $4.9m{s^{ - 2}}$
B) $9.8m{s^{ - 2}}$
C) $19.8m{s^{ - 2}}$
D) None of the above
Answer
232.8k+ views
Hint: Think about Newton's law of gravitation and try to use the relation between acceleration due to gravity and the universal gravitational constant. The acceleration due to gravity depends on the radius of the planet or satellite and so that has different values for moon and earth as their radii vary.
Complete step by step solution:
The law of universal gravity of Newton is normally that any particle attracts every other particle in the universe with a force strictly proportional and inversely proportional to the distance between its cores. Or, the rule says that every point mass is attracted by force along the line between the two lines, by every other point mass. The power is proportional to the production of the two masses and inversely proportional to their distance.
Now, according to gravity equation, we know that
\[g \propto \dfrac{{GM}}{{{R^2}}}\]
$g$ is acceleration due to gravity,
$G$ is the gravitational constant,
$M$ is the mass of the body and,
$R$ is the radius of the body.
Now, the acceleration due to gravity ${g_p} $ on different planet can be written as
${g_p} = (\dfrac{{{M_p}}}{{{M_e}}} \times {(\frac{{{R_e}}}{{{R_p}}})^2})g$
${M_p} $ is the mass of the object on planet
${M_e} $ is the mass of the object on earth.
${R_p} $ is the radius of the object on the planet.
${R_e} $ is the radius of the object on earth.
Putting the values given in the question,
$ \Rightarrow {g_p} = 2 \times {(\dfrac{1}{2})^2}g$
${g_p} = \dfrac{g}{2}$
${g_p} = 4.9m{s^ {- 2}} $
The acceleration due to gravity ${g_p} $ on different planets is $4.9m{s^ {- 2}} $.
Hence the correct option is A.
Note: Since the force of gravity is directly commensurate with the mass of all interacting objects, more large objects are drawn with a greater force. If the mass of all objects increases, the gravitational force also increases between them. If the mass of one object is doubled, the gravitational power is doubled between them. If the mass of one object is multiplied, the force of gravity is tripled between the objects. The force of gravity on these two objects is quadrupled by the density of all objects, etc.
As the gravitational power is inversely equal to the dividing distance square between the two interacting objects, further dividing distance produces weaker gravitational force. The force of gravitational interaction between them also reduces, when two objects are isolated from each other.
Complete step by step solution:
The law of universal gravity of Newton is normally that any particle attracts every other particle in the universe with a force strictly proportional and inversely proportional to the distance between its cores. Or, the rule says that every point mass is attracted by force along the line between the two lines, by every other point mass. The power is proportional to the production of the two masses and inversely proportional to their distance.
Now, according to gravity equation, we know that
\[g \propto \dfrac{{GM}}{{{R^2}}}\]
$g$ is acceleration due to gravity,
$G$ is the gravitational constant,
$M$ is the mass of the body and,
$R$ is the radius of the body.
Now, the acceleration due to gravity ${g_p} $ on different planet can be written as
${g_p} = (\dfrac{{{M_p}}}{{{M_e}}} \times {(\frac{{{R_e}}}{{{R_p}}})^2})g$
${M_p} $ is the mass of the object on planet
${M_e} $ is the mass of the object on earth.
${R_p} $ is the radius of the object on the planet.
${R_e} $ is the radius of the object on earth.
Putting the values given in the question,
$ \Rightarrow {g_p} = 2 \times {(\dfrac{1}{2})^2}g$
${g_p} = \dfrac{g}{2}$
${g_p} = 4.9m{s^ {- 2}} $
The acceleration due to gravity ${g_p} $ on different planets is $4.9m{s^ {- 2}} $.
Hence the correct option is A.
Note: Since the force of gravity is directly commensurate with the mass of all interacting objects, more large objects are drawn with a greater force. If the mass of all objects increases, the gravitational force also increases between them. If the mass of one object is doubled, the gravitational power is doubled between them. If the mass of one object is multiplied, the force of gravity is tripled between the objects. The force of gravity on these two objects is quadrupled by the density of all objects, etc.
As the gravitational power is inversely equal to the dividing distance square between the two interacting objects, further dividing distance produces weaker gravitational force. The force of gravitational interaction between them also reduces, when two objects are isolated from each other.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

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

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Understanding Uniform Acceleration in Physics

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

