# Given: G = 6.67259 x 10^{-11} N x m^2/kg^2 A satellite moves in a circular orbit around the...

## Question:

Given: G = 6.67259 x 10{eq}^{-11} N x m^2/kg^2 {/eq}

A satellite moves in a circular orbit around the earth at a speed of 2090 m/s.

A. Find the altitude above the surface of the earth. Answer in units of m.

B. Find the period of the satellite's orbit. Answer in units of h.

## Gravity:

The tendency of every matter and objects to impose an attractive force on all other objects is termed as gravity. As the gap between objects increases, the gravity effect diminishes. The gravity effect is a function of masses of objects.

Given Data

• The gravitational constant is: {eq}G = 6.67259 \times {10^{ - 11}}\;{\rm{N}} \cdot {{\rm{m}}^2}{\rm{/k}}{{\rm{g}}^{\rm{2}}} {/eq}.
• The speed of the satellite is: {eq}v = 2090\;{\rm{m/s}} {/eq}.

(A)

The formula to calculate the height of the satellite from the center of the Earth is given by,

{eq}\begin{align*} {v^2} &= \dfrac{{Gm}}{x}\\ x &= \dfrac{{Gm}}{{{v^2}}} \end{align*} {/eq}

Here, {eq}m {/eq} is the mass of the Earth, having a standard value of {eq}5.97 \times {10^{24}}\;{\rm{kg}}. {/eq}

The formula to calculate the satellitel's altitude from the Earth's surface is given by,

{eq}A = x - r {/eq}

Here, {eq}r {/eq} is the radius of the Earth, having a standard value of {eq}6.4 \times {10^6}\;{\rm{m}} {/eq}.

Substitute the formula of the satellite's height from the Earth's center in above formula.

{eq}A = \dfrac{{Gm}}{{{v^2}}} - r {/eq}

Substitute all the values in the above formula.

{eq}\begin{align*} A &= \dfrac{{6.67259 \times {{10}^{ - 11}} \times 5.97 \times {{10}^{24}}}}{{{{2090}^2}}} - 6.4 \times {10^6}\\ & = 8.479 \times {10^7}\;{\rm{m}}\\ & \approx 8.48 \times {10^7}\;{\rm{m}} \end{align*} {/eq}

Thus, the satellite altitude above the surface of the Earth is {eq}8.48 \times {10^7}\;{\rm{m}} {/eq}.

(B)

The expression to calculate the period of the satellite's orbit is given by,

{eq}T = 2\pi \times \dfrac{x}{v} {/eq}

Substitute the formula of the satellite's height from the Earth's center in the above formula.

{eq}\begin{align*} T &= 2\pi \times \dfrac{{\left( {\dfrac{{Gm}}{{{v^2}}}} \right)}}{v}\\ & = 2\pi \times \dfrac{{Gm}}{{{v^3}}} \end{align*} {/eq}

Substitute all the values in the above formula.

{eq}\begin{align*} T &= 2\pi \times \dfrac{{6.67259 \times {{10}^{ - 11}} \times 5.97 \times {{10}^{24}}}}{{{{2090}^3}}}\\ & = 2.741 \times {10^5}\;{\rm{s}}\\ & = 2.741 \times {10^5}\;{\rm{s}} \times \left( {\dfrac{{1\;{\rm{min}}}}{{60\;{\rm{s}}}}} \right) \times \left( {\dfrac{{1\;{\rm{h}}}}{{60\;{\rm{min}}}}} \right)\\ & = 76.13\;{\rm{h}} \end{align*} {/eq}

Thus, the period of the satellite's orbit is {eq}76.13\;{\rm{h}} {/eq}.

Gravitational Pull of the Earth: Definition & Overview

from

Chapter 15 / Lesson 17
31K

Earth's gravitational pull is often misunderstood, but without it, life on Earth would be impossible. In this lesson, we'll define the gravitational pull and give some examples of how it is used. A quiz is provided to test your understanding.