# 13.3 Gravitational potential energy and total energy  (Page 2/6)

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## Lifting a payload

How much energy is required to lift the 9000-kg Soyuz vehicle from Earth’s surface to the height of the ISS, 400 km above the surface?

## Strategy

Use [link] to find the change in potential energy of the payload. That amount of work or energy must be supplied to lift the payload.

## Solution

Paying attention to the fact that we start at Earth’s surface and end at 400 km above the surface, the change in U is

$\text{Δ}U={U}_{\text{orbit}}-{U}_{\text{Earth}}=-\frac{G{M}_{\text{E}}m}{{R}_{\text{E}}+\phantom{\rule{0.2em}{0ex}}400\phantom{\rule{0.2em}{0ex}}\text{km}}-\left(-\frac{G{M}_{\text{E}}m}{{R}_{\text{E}}}\right).$

We insert the values

$m=9000\phantom{\rule{0.2em}{0ex}}\text{kg,}\phantom{\rule{1em}{0ex}}{M}_{\text{E}}=5.96\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{24}\text{kg,}\phantom{\rule{1em}{0ex}}{R}_{\text{E}}=6.37\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}\text{m}$

and convert 400 km into $4.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{m}$ . We find $\text{Δ}U=3.32\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{10}\phantom{\rule{0.2em}{0ex}}\text{J}$ . It is positive, indicating an increase in potential energy, as we would expect.

## Significance

For perspective, consider that the average US household energy use in 2013 was 909 kWh per month. That is energy of

$909\phantom{\rule{0.2em}{0ex}}\text{kWh}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}1000\phantom{\rule{0.2em}{0ex}}\text{W/kW}\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}3600\phantom{\rule{0.2em}{0ex}}\text{s/h}=3.27\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{9}\phantom{\rule{0.2em}{0ex}}\text{J per month.}$

So our result is an energy expenditure equivalent to 10 months. But this is just the energy needed to raise the payload 400 km. If we want the Soyuz to be in orbit so it can rendezvous with the ISS and not just fall back to Earth, it needs a lot of kinetic energy. As we see in the next section, that kinetic energy is about five times that of $\text{Δ}U$ . In addition, far more energy is expended lifting the propulsion system itself. Space travel is not cheap.

Check Your Understanding Why not use the simpler expression $\text{Δ}U=mg\left({y}_{2}-{y}_{1}\right)$ ? How significant would the error be? (Recall the previous result, in [link] , that the value g at 400 km above the Earth is $8.67\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}$ .)

The value of g drops by about 10% over this change in height. So $\text{Δ}U=mg\left({y}_{2}-{y}_{1}\right)$ will give too large a value. If we use $g=9.80\phantom{\rule{0.2em}{0ex}}\text{m/s}$ , then we get

$\text{Δ}U=mg\left({y}_{2}-{y}_{1}\right)=3.53\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{10}\phantom{\rule{0.2em}{0ex}}\text{J}$

which is about 6% greater than that found with the correct method.

## Conservation of energy

In Potential Energy and Conservation of Energy , we described how to apply conservation of energy for systems with conservative forces. We were able to solve many problems, particularly those involving gravity, more simply using conservation of energy. Those principles and problem-solving strategies apply equally well here. The only change is to place the new expression for potential energy into the conservation of energy equation, $E={K}_{1}+{U}_{1}={K}_{2}+{U}_{2}$ .

$\frac{1}{2}\phantom{\rule{0.1em}{0ex}}m{v}_{1}^{2}-\frac{GMm}{{r}_{1}}=\frac{1}{2}\phantom{\rule{0.1em}{0ex}}m{v}_{2}^{2}-\frac{GMm}{{r}_{2}}$

Note that we use M , rather than ${M}_{\text{E}}$ , as a reminder that we are not restricted to problems involving Earth. However, we still assume that $m\text{<}\phantom{\rule{0.2em}{0ex}}\text{<}M$ . (For problems in which this is not true, we need to include the kinetic energy of both masses and use conservation of momentum to relate the velocities to each other. But the principle remains the same.)

## Escape velocity

Escape velocity is often defined to be the minimum initial velocity of an object that is required to escape the surface of a planet (or any large body like a moon) and never return. As usual, we assume no energy lost to an atmosphere, should there be any.

Consider the case where an object is launched from the surface of a planet with an initial velocity directed away from the planet. With the minimum velocity needed to escape, the object would just come to rest infinitely far away, that is, the object gives up the last of its kinetic energy just as it reaches infinity, where the force of gravity becomes zero. Since $U\to 0\phantom{\rule{0.2em}{0ex}}\text{as}\phantom{\rule{0.2em}{0ex}}r\to \infty$ , this means the total energy is zero. Thus, we find the escape velocity    from the surface of an astronomical body of mass M and radius R by setting the total energy equal to zero. At the surface of the body, the object is located at ${r}_{1}=R$ and it has escape velocity ${v}_{1}={v}_{\text{esc}}$ . It reaches ${r}_{2}=\infty$ with velocity ${v}_{2}=0$ . Substituting into [link] , we have

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