Question

A satellite revolves around a planet at an altitude equal to the radius of the planet. the force of gravitational interaction between the satellite and the planet is f0. then the satellite moves to a different orbit, so that its altitude is tripled. find the new force of gravitational interaction f2.

Answer 1

We can use the equation for the force of gravity to solve this question. F = G M1 M2 / R^2 Let R be the radius of the planet. f0 = G M1 M2 / (2R)^2 f0 = (1/4) x (G M1 M2 / R^2) We can find an expression for f2. f2 = G M1 M2 / (4R)^2 f2 = (1/16) x (G M1 M2 / R^2) We can express f2 in terms of f0: f2 = (4/16) x f0 f2 = 0.25 f0

Answer 2

The new force of gravitational interaction F₂ is ¹/₄ F₀

$\texttt{ }$

Further explanation

Newton's gravitational law states that the force of attraction between two objects can be formulated as follows:

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

F = Gravitational Force ( Newton )

G = Gravitational Constant ( 6.67 × 10⁻¹¹ Nm² / kg² )

m = Object's Mass ( kg )

R = Distance Between Objects ( m )

Let us now tackle the problem !

$\texttt{ }$

Given:

initial altitude of the satellite = h₁ = R

initial force of gravitational interaction = F₀

final altitude of the satellite = h₂ = 3R

radius of the planet = R

Asked:

final force of gravitational interaction = F₂

Solution:

$F_o : F_2 = G \frac{ M m} { (R + h_1)^2 } : G \frac{ M m} { (R + h_2)^2 }$

$F_o : F_2 = \frac{ 1 } { (R + h_1)^2 } : \frac{ 1 } { (R + h_2)^2 }$

$F_o : F_2 = (R + h_2)^2 : (R + h_1)^2$

$F_o : F_2 = (R + 3R)^2 : (R + R)^2$

$F_o : F_2 = (4R)^2 : (2R)^2$

$F_o : F_2 = 4^2 : 2^2$

$F_o : F_2 = 16 : 4$

$F_o : F_2 = 4 : 1$

$\boxed {F_2 = \frac{1}{4} F_o}$

$\texttt{ }$

Learn more

• Impacts of Gravity : brainly.com/question/5330244
• Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
• The Acceleration Due To Gravity : brainly.com/question/4189441

$\texttt{ }$

Answer details

Grade: High School

Subject: Physics

Chapter: Gravitational Fields