
What Is Acceleration Due To Gravity On Jupiter?
Answer
162.9k+ 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
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

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

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Charging and Discharging of Capacitor

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

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
