# From the earth's surface, a rocket is fired horizontally. What must be this initial velocity if...

## Question:

From the earth's surface, a rocket is fired horizontally. What must be this initial velocity if the rocket is to orbit at a radius of 3.32E07 m from the center of the earth? Hint: You must use two conservation laws.

## A practice problem on horizontally launched satellite:

A satellite is an object in space that goes around another (usually larger) object. The repeated path followed by the satellite around its primary object is called the orbit. There are natural satellites like the Moon to the Earth and artificial satellites sent into space to orbit around any celestial body for various purposes. An artificial satellite is made to travel into space from the surface of the Earth. The initiation of the process to send the satellite to an orbit is called the launching. Once launched, an artificial satellite to the Earth is made to reach the desired radial distance from the Earth to start orbiting.

A satellite indeed is a projectile that always falls towards the earth but never hits the Earth. This interesting happening is due to a combination of two reasons. One is that the Earth is round. Starting at a point on the surface of the Earth, if we travel a horizontal distance, there will be a corresponding vertical separation due to the Earth's curvature. This would not have happened if the surface of the earth is flat. The second reason is that any object thrown into the air has to fall down through a fixed distance per second due to gravity at a location. If the fall of the projectile due to gravity in a given time is made equal to or greater than the vertical separation due to curvature and horizontal distance covered at the same time, then the projectile never hits the Earth but goes around the Earth though it accelerates towards the Earth. The present problem is concerned with finding the initial velocity of a horizontally projected satellite to orbit the earth at a defined height.

## Answer and Explanation:

A horizontally projected object has to fall towards the Earth due to gravity. The vertical distance of the gravity fall in a time interval t is given by, {eq}S_v=\frac{1}{2}gt^2 {/eq}. Near the surface of the Earth, the vertical downward distance covered by the projected object in the first second is {eq}\frac {1}{2}g\ (\text{= say approximately 5 m }) {/eq} due to gravity. Due to the curvature of the Earth, the vertical separation of any object traveling horizontally along the horizon is about 5 m for every 8000 m of horizontal distance covered. Therefore a projected object traveling at a horizontal speed greater than 8000 m/sec, due to gravity it falls a vertical distance less than 5 m for a horizontal journey of 8000 m. Hence it never hits the Earth. This is the condition that a projectile turns to a satellite.

A horizontally launched satellite circling the Earth at a radius r ( {eq}= 3.32 \times 10^7\ m ) {/eq} measured from the center of the Earth is considered in the problem. The solution is to find the value of the initial horizontal velocity. The approach to finding the solution is to equate the kinetic energy initially imparted to the satellite to the sum of the increase in potential energy {eq}\Delta PE {/eq} and the energy of motion {eq}\frac{1}{2} mV_O^2 {/eq} of the satellite in orbit, here m is the mass of the satellite and {eq}V_O {/eq} is the velocity of a satellite in orbit. Thus the initial horizontal velocity of the satellite can be found from energy balance. Here indirectly the conservative law for energy (both kinetic and potential) is used. Air resistance is ignored in calculations.

#### {eq}\text {The difference in potential energy } \Delta PE\ \text{ of the satellite with its positions at the surface of the Earth and the orbit} : {/eq}

To lift any object against the attractive force of gravity, work needs to be done. The work done to move an object to raise its height is stored as potential energy. So the difference in potential energy of the satellite before launch on the surface of the Earth with average radius R ({eq}= 6.371 \times 10^6\ m ) {/eq} and the potential energy at the orbital radius r ({eq}=3.32 \times 10^7\ m, \text{ measured from the center of the Earth } ) {/eq} is the amount of energy required just to relocate the satellite from the surface of the Earth to the orbit.

The gravitational potential energy of the satellite on the surface of the Earth; at the average radius of the Earth R ({eq}= 6.371 \times 10^6\ m{/eq} ):

{eq}PE_S \displaystyle =- \frac{GMm}{R} \qquad \cdots(1) \qquad ( G= 6.67408 \times 10^{-11}\ m^3kg^{-1}s^{-2}\ \text{ is the universal gravitation constant, } M= 5.972 \times 10^{24}\ kg, \text{is the mass of the Earth, and m is the mass of the satellite} ) {/eq}

The gravitational potential energy of the satellite, at the orbital radius ({eq}r = 3.32 \times 10^7\ m;{/eq} the radius measured from the center of the Earth ) :

{eq}PE_O \displaystyle =- \frac{GMm}{r} \qquad \cdots(2)\\ \text{The difference in gravitational potential energy in re-locating the satellite from surface of the Eath to the orbit:}\\ \Delta PE= PE_O-PE_S \displaystyle = -\frac{GMm}{r} +\frac{GMm}{R} \\ \hspace 21 pt \displaystyle =\frac{GMm}{R}-\frac{GMm}{r} \\ \Delta PE \displaystyle =GMm \left \{ \frac{1}{R}-\frac{1}{r} \right \} \qquad \cdots(3) {/eq}

#### {eq}\text {The kinetic energy of the orbiting satellite at radial distance r (=3.32 } \times 10^7\ m\ ) : {/eq}

In order for the satellite to sustain orbiting the earth, the centrifugal force of the satellite and the force of gravity are to be equal.

The force of gravity at r, the orbital radius:

{eq}\displaystyle F_g=\frac{GMm}{ r^2} \qquad \cdots(4) {/eq}

The centrifugal force of the satellite orbiting at r ( {eq}=3.32 \times 10^7\ m\ ) :{/eq}

{eq}\displaystyle F_c=\frac{mV_O^2}{ r} \qquad \cdots(5)\\ \text{Equating the force due to gravity and the centrifugal force given by equations 4 & 5: }\\ \frac{GMm}{ r^2} \displaystyle =\frac{mV_O^2}{ r} \qquad \cdots(6)\\ \Rightarrow \displaystyle V_O=\sqrt{ \frac{GM}{r}}\\ \text{The kinetic energy of the orbiting satellite is:}\\ \displaystyle KE_O=\frac{1}{2}mV_O^2 \qquad ( V_O\ \text{ is the velocity of the satellite in the orbit} )\\ \text{The initial horizontal velocity has to be such that total initial } KE_{i Total} \text{ should be equal to the required increase of the gravitational potential energy plus the orbital kinetic enery}\\ \displaystyle KE_{i Total}=KE_O+ \Delta PE\\ \frac{1}{2}mV_{IH}^2= \frac{1}{2}mV_O^2+GMm \left \{ \frac{1}{R}-\frac{1}{r} \right \}\\ V_{IH}=\sqrt { V_O^2+2GM \left \{ \frac{1}{R}-\frac{1}{r} \right \} }\\ \hspace 18 pt = \sqrt { \frac{GM}{r}+2GM \left \{ \frac{1}{R}-\frac{1}{r} \right \} } \\ \hspace 18 pt = \sqrt { GM \left \{ \frac{1}{r}+2 \left ( \frac{1}{R}-\frac{1}{r} \right ) \right \} } \\ \hspace 18 pt = \sqrt { GM \left \{ \frac{2}{R}-\frac{1}{r} \right \} } \qquad \cdots(7)\\ \text{Substituting the values for the universal gravitation constant } G= 6.67408 \times 10^{-11}\ m^3kg^{-1}s^{-2},\ \text{mass of the Earth } M= 5.972 \times 10^{24}\ kg, \text{ average radius of the Earth } R=6.371 \times 10^6\ m,\\ \text{and given radius of orbit } r\ = 3.32\ \times 10^7\ m\ \text{in equation-7}\\ \hspace 18 pt \displaystyle = \sqrt { 6.67408 \times 10^{-11} \times 5.972 \times 10^{24} \times \left \{ \frac{2}{6.371 \times 10^6}-\frac{1}{3.32 \times 10^7} \right \} }\\ \hspace 18 pt=10.635632 \times^3\ m/sec\ or\ 10.635632\ km/sec {/eq}

{eq}\textbf{ The initial velocity } V_{IH} \textbf{ of a horizontally launched satellite from surface of the Earth to the orbit at } r=3.32 \times 10^7\ m\ \textbf{from center of the Earth, } V_{IH}=10.635632\ km/sec. {/eq}

#### Learn more about this topic:

How Orbits Are Influenced by Gravity & Energy

from Basics of Astronomy

Chapter 25 / Lesson 6
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