What Is Acceleration Due To Gravity On Jupiter?
Answer
298.5k+ views
Hint:The universal gravitational constant is related to the attractive gravitational force between two bodies separated by a distance r. The acceleration due to gravity on earth is the acceleration experienced by anybody during free fall due to the attractive gravitational force of the earth's surface. It is not a universal constant. On earth, it is usually taken to be \[9.8{\rm{ }}m{s^{ - 2}}\].
Formula used:
Gravitational Force,
\[F = GMm/{r^2}\]
Where G = universal gravitational constant.
M= mass of the planet (earth)
m = mass of the lighter object
r = distance between the two objects
Acceleration due to gravity,
\[g = GM/{r^2}\]
where M= mass of the planet (earth)
r = radius of the planet
Complete step by step solution:
Given: mass of Jupiter = 319 times the mass of earth
Radius of Jupiter = 11.2 times the radius of earth
From Newton’s law of gravitation,
Force, \[F = GMm/{r^2}\]----- (1)
where G= universal gravitational constant = \[6.674 \times {10^{ - 11}}{m^3}k{g^{ - 1}}{s^{ - 2}}\]
Also, from Newton’s third law of motion,
\[F = mg\]------ (2)
where g = acceleration due to gravity on earth = \[9.8{\rm{ }}m{s^{ - 2}}\]
Equating (1) and (2)
\[g = GM/{r^2}\]------ (3)
Let \[\]\[M\], \[r\], \[{M_j}\] and \[{r_j}\] be masses and radii of earth and Jupiter respectively. Then according to the question,
\[{M_j} = 319M\]--(4) and \[{r_j} = 11.2r\]--- (5)
Using equation (3) to calculate \[{g_j}\]= acceleration due to gravity on Jupiter,
\[{g_j} = G{M_j}/{r^2}_j\]---- (6)
Substituting equations (4) and (5) in (6), we get,
\[{g_j} = G \times 319M/{(11.2r)^2}\]
\[\Rightarrow{g_j} = 2.54g\]
\[\Rightarrow {g_j} = 2.54 \times 9.8\]
\[\therefore {g_j} = 24.58\,m{s^{ - 2}}\]
Hence the acceleration due to gravity on Jupiter is \[{g_j} = 24.58\,m{s^{ - 2}}\].
Note: Although acceleration due to gravity is a constant, it is not a universal constant like universal gravitational constant, G. It varies on earth with change in reference surface such as when measured on a mountain or in the depths of water bodies like seas and oceans.
Formula used:
Gravitational Force,
\[F = GMm/{r^2}\]
Where G = universal gravitational constant.
M= mass of the planet (earth)
m = mass of the lighter object
r = distance between the two objects
Acceleration due to gravity,
\[g = GM/{r^2}\]
where M= mass of the planet (earth)
r = radius of the planet
Complete step by step solution:
Given: mass of Jupiter = 319 times the mass of earth
Radius of Jupiter = 11.2 times the radius of earth
From Newton’s law of gravitation,
Force, \[F = GMm/{r^2}\]----- (1)
where G= universal gravitational constant = \[6.674 \times {10^{ - 11}}{m^3}k{g^{ - 1}}{s^{ - 2}}\]
Also, from Newton’s third law of motion,
\[F = mg\]------ (2)
where g = acceleration due to gravity on earth = \[9.8{\rm{ }}m{s^{ - 2}}\]
Equating (1) and (2)
\[g = GM/{r^2}\]------ (3)
Let \[\]\[M\], \[r\], \[{M_j}\] and \[{r_j}\] be masses and radii of earth and Jupiter respectively. Then according to the question,
\[{M_j} = 319M\]--(4) and \[{r_j} = 11.2r\]--- (5)
Using equation (3) to calculate \[{g_j}\]= acceleration due to gravity on Jupiter,
\[{g_j} = G{M_j}/{r^2}_j\]---- (6)
Substituting equations (4) and (5) in (6), we get,
\[{g_j} = G \times 319M/{(11.2r)^2}\]
\[\Rightarrow{g_j} = 2.54g\]
\[\Rightarrow {g_j} = 2.54 \times 9.8\]
\[\therefore {g_j} = 24.58\,m{s^{ - 2}}\]
Hence the acceleration due to gravity on Jupiter is \[{g_j} = 24.58\,m{s^{ - 2}}\].
Note: Although acceleration due to gravity is a constant, it is not a universal constant like universal gravitational constant, G. It varies on earth with change in reference surface such as when measured on a mountain or in the depths of water bodies like seas and oceans.
Recently Updated Pages
Four persons A B C and D initially at the corners of class 11 physics JEE_Main

What is the difference between Conduction and conv class 11 physics JEE_Main

Moment of inertia of solid sphere about its diameter class 11 physics JEE_Main

If a piece of ice floating on the surface of water class 11 physics JEE_Main

At what temperature speed of sound in air will be doubled class 11 physics JEE_Main

A closed organ pipe and an open organ pipe are tuned class 11 physics JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2026-27 PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 2 Motion In A Straight Line - 2026-27 Free PDF Download (Login Required)

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2026-27 Free PDF Download (Sign-in Required)

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

CBSE Notes Class 11 Physics Chapter 2 - Motion in a Straight Line - 2026-27 PDF Download (Login Required)

