if the density of earth increases by 20% and radius decreases by 20% t...
G = GM / R^2
= G � (4/3 � π � R^3 � d) / R^2
= 4πGdR/3
From above
g ∝ dR
g^1/g^2= (d₁/d₂) � (R₁/R₂)
g^1/g^2 = (100 / 120) � (100 / 80)
g^1/g^2 = 25/24
g^2 = 24g^1/25
= 0.96 g^1
= 0.96 � 9.8 m/s^2
= 9.408 m/s^2
Value of g on surface of earth will be 9.408 m/s^2
This question is part of UPSC exam. View all Class 9 courses
if the density of earth increases by 20% and radius decreases by 20% t...
Explanation:
To determine the new value of "g" on the surface of the Earth, we need to consider the effects of both the increase in density and the decrease in radius. Let's break it down step by step:
Step 1: Increase in Density
When the density of the Earth increases by 20%, it means that the mass of the Earth increases while the volume remains the same. As density is defined as mass divided by volume, an increase in mass with constant volume leads to an increase in density.
Step 2: Decrease in Radius
When the radius of the Earth decreases by 20%, it means that the distance from the center of the Earth to its surface decreases. This reduction in radius affects the gravitational force experienced on the surface of the Earth.
Step 3: Calculation of New "g" Value
To calculate the new value of "g" on the surface of the Earth, we can use the formula for gravitational acceleration:
g = (G * M) / R^2
where:
- G is the gravitational constant (approximately 6.674 x 10^-11 N*m^2/kg^2)
- M is the mass of the Earth
- R is the radius of the Earth
Since the mass of the Earth has increased due to the increase in density, we can denote it as M'. Similarly, since the radius of the Earth has decreased, we can denote it as R'. Substituting these values into the formula, we get:
g' = (G * M') / (R')^2
Step 4: Combining the Effects
We need to consider the combined effect of both the increase in density and the decrease in radius. Let's assume the original values of density, mass, and radius are denoted as ρ, M, and R, respectively.
The new mass of the Earth, M', can be calculated as:
M' = M * (1 + 20%) = M * 1.2
The new radius of the Earth, R', can be calculated as:
R' = R * (1 - 20%) = R * 0.8
Substituting these values into the formula, we get:
g' = (G * M') / (R')^2
g' = (G * M * 1.2) / (R * 0.8)^2
g' = (1.2 * G * M) / (0.64 * R^2)
g' = 1.875 * (G * M) / R^2
Step 5: Conclusion
The new value of "g" on the surface of the Earth, denoted as g', is approximately 1.875 times the original value of "g". Therefore, the gravitational acceleration on the surface of the Earth would increase by approximately 87.5% if the density of the Earth increases by 20% and the radius decreases by 20%.
To make sure you are not studying endlessly, EduRev has designed Class 9 study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in Class 9.