Chapter 6.6, Problem 27E

(a).

To determine

The work needed to move a body between two points under the influence of gravitational force of other body.

Answer to Problem 27E

The work needed to move the second body from $r=a$ to $r=b$ is given by the expression $G{m}_1{m}_2\left(\frac{1}{a}-\frac{1}{b}\right)$ .

Explanation of Solution

Given: Gravitational force of attraction between two bodies of masses $m_1$ , $m_2$ and distance of separation $r$ is $F\left(r\right)=G\frac{m_1{m}_2}{r^2}$ , where $G$ is the gravitational constant.

The initial position of the object is $r=a$ and the final position is $r=b$ .

Concept Used:

The work done by a force $F$ in moving a particle from $x_1$ to $x_2$ is given by

  $W={{\displaystyle \int }}_{x_1}^{x_2}F\left(x\right)dx$

Put the expression for the gravitational force $F\left(r\right)$ and the corresponding limits of $r$ in the above work equation to write

  $W_{a\to b}={{\displaystyle \int }}_a^{b}G\frac{m_1{m}_2}{r^2}dr$

  $={-G\frac{m_1{m}_2}{r}]}_a^b=G{m}_1{m}_2\left(\frac{1}{a}-\frac{1}{b}\right)$

Therefore, the work needed to move the second body from $r=a$ to $r=b$ is given by the expression

  $W_{a\to b}=G{m}_1{m}_2\left(\frac{1}{a}-\frac{1}{b}\right)$

(b).

To determine

To Calculate: The work required to launch a satellite from the earth to an orbit of certain radius.

Answer to Problem 27E

The work needed to put the satellite into orbit is $8.5\times {10}^{9}J$ .

Explanation of Solution

Given:

• Mass of the satellite is $1000kg$ .
• Mass of the earth is $5.98\times {10}^{24}kg$ .
• Radius of the earth is $6.37\times {10}^{6}m$ .
• The orbit is $1000km$ high from the earth surface.
• The value of $G$ is $6.67\times {10}^{-11}N\cdot m^2/k{g}^2$ .

Calculation: The radius of the orbit is the sum of radius of the earth and the orbit height. So

  $r_{orbit}=6.37\times {10}^{6}m+1000km$

  $=6.37\times {10}^{6}m+{10}^{6}m$

  $=7.37\times {10}^{6}m$

In the expression for work $W_{a\to b}$ derived in the previous subpart, put

• mass of the earth for $m_1$
• mass of the satellite for $m_2$
• radius of the earth for $a$ and
• radius of the orbit $r_{orbit}$ for $b$ so that

  $W=\left(6.67\times {10}^{-11}\right)\cdot \left(5.98\times {10}^{24}\right)\times {10}^3\times \left(\frac{1}{6.37\times {10}^6}-\frac{1}{7.37\times {10}^6}\right)$

  $=6.67\times 5.98\times \left(\frac{1}{6.37}-\frac{1}{7.37}\right)\times {10}^{\left(-11+24+3-6\right)}$

  $=0.85\times {10}^{10}$

Therefore, the work needed to put the satellite into orbit is $8.5\times 10^{9}J$ .