Respuesta :
Gravitational force on a satellite is given by the formula
[tex]F = \frac{GMm}{r^2}[/tex]
now here we know that force on the satellite is F when its distance from center of Earth is R
Now the distance from the center of earth will be 3R so the force is given as
[tex]F' = \frac{GMm}{(3R)^2}[/tex]
[tex]F' = \frac{GMm}{9R^2}[/tex]
so if we compare it with initial value of force then it is
[tex]F' = \frac{F}{9}[/tex]
so correct answer is
[tex](1) F/9[/tex]
The gravitational force on the satellite is (1) F/9
[tex]\texttt{ }[/tex]
Further explanation
Let's recall the Gravitational Force formula:
[tex]\boxed {F = G\ \frac{m_1 m_2}{R^2}}[/tex]
where:
F = Gravitational Force ( N )
G = Gravitational Constant ( = 6.67 × 10⁻¹¹ Nm²/kg² )
m = mass of object ( kg )
R = distance between object ( m )
Let us now tackle the problem!
[tex]\texttt{ }[/tex]
Given:
initial distance of the satellite from the centre of the Earth = R₁ = R
initial force of gravity on the satellite = F₁ = F
final distance of the satellite from the centre of the Earth = R₂ = 3R
Asked:
initial force of gravity on the satellite = F₂ = ?
Solution:
[tex]F_2 : F_1 = G\ \frac{M m}{(R_2)^2} : G\ \frac{M m}{(R_1)^2}[/tex]
[tex]F_2 : F_1 = \frac{1}{(R_2)^2} : \frac{1}{(R_1)^2}[/tex]
[tex]F_2 : F_1 = (R_1)^2} : (R_2)^2[/tex]
[tex]F_2 : F = R^2 : (3R)^2[/tex]
[tex]F_2 : F = R^2 : 9R^2[/tex]
[tex]F_2 : F = 1 : 9[/tex]
[tex]\boxed {F_2 = \frac{1}{9} F}[/tex]
[tex]\texttt{ }[/tex]
Learn more
- Unit of G : https://brainly.com/question/1724648
- Velocity of Runner : https://brainly.com/question/3813437
- Kinetic Energy : https://brainly.com/question/692781
- Acceleration : https://brainly.com/question/2283922
- The Speed of Car : https://brainly.com/question/568302
[tex]\texttt{ }[/tex]
Answer details
Grade: High School
Subject: Mathematics
Chapter: Gravitational Force
