Question

When a satellite is a distance R from the center of the Earth, the force of gravity on the satellite is F. What is the gravitational force on the satellite when it’s distance from the center of the Earth is 3R?
(1) F/9
(2) F/3
(3) F
(4) 9F

Answer 1

Gravitational force on a satellite is given by the formula

$F = \frac{GMm}{r^2}$

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

$F' = \frac{GMm}{(3R)^2}$

$F' = \frac{GMm}{9R^2}$

so if we compare it with initial value of force then it is

$F' = \frac{F}{9}$

so correct answer is

$(1) F/9$

Answer 2

The gravitational force on the satellite is (1) F/9

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Further explanation

Let's recall the Gravitational Force formula:

$\boxed {F = G\ \frac{m_1 m_2}{R^2}}$

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!

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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:

$F_2 : F_1 = G\ \frac{M m}{(R_2)^2} : G\ \frac{M m}{(R_1)^2}$

$F_2 : F_1 = \frac{1}{(R_2)^2} : \frac{1}{(R_1)^2}$

$F_2 : F_1 = (R_1)^2} : (R_2)^2$

$F_2 : F = R^2 : (3R)^2$

$F_2 : F = R^2 : 9R^2$

$F_2 : F = 1 : 9$

$\boxed {F_2 = \frac{1}{9} F}$

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Learn more

• Unit of G : brainly.com/question/1724648
• Velocity of Runner : brainly.com/question/3813437

• Kinetic Energy : brainly.com/question/692781

• Acceleration : brainly.com/question/2283922

• The Speed of Car : brainly.com/question/568302

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Answer details

Grade: High School

Subject: Mathematics

Chapter: Gravitational Force