# Total Energy of the ISS orbiting the Earth [duplicate]

I had a bit of confusion earlier about a discussion in astrophysics.

I was reading that the ISS orbits the earth at $8 kms^{-1}$ but then when I calculated the total energy that the ISS had, I was considering:

1. It's Kinetic Energy $\frac{1}{2}mv^2$
2. It's Gravitational Potential Energy $-\frac{GMm}{r}$

where

• $m$ is the mass of the ISS
• $M$ is the mass of the Earth
• $v$ is the speed of the ISS

When I did my calculations, something got me confused.

Using the following data:

ISS circles the Earth at a height of $4.0 \times 10^5$m.
ISS has a mass of $4.2 \times 10^5$ kg
The radius of the Earth is $6.4 \times 10^6$ m
The mass of the Earth is $6.0 \times 10^24$kg

It turned out the magnitude of the Gravitational Potential Energy is greater than the magnitude of the Kinetic energy.

$$E_K < |E_{GPE}|$$

But in the lecture notes I was reading for astrophysics, the overall Energy is negative as you add both Energies together.

Is it possible for the energy of an object to be negative, and is it valid in this case?

I was thinking to add the magnitudes of both $E_K$ and $E_{GPE}$ as I assumed the total energy of the ISS was equal to the sum of these energies, hence giving a positive value.

## marked as duplicate by sammy gerbil, David Hammen, stafusa, Bill N, Jon CusterJun 4 '18 at 20:43

The sum of the potential energy and the kinetic energy is denoted by $$E = T + V$$ Where $T$ and $V$ are the kinetic and potential energy respectively. $T$ and $V$ are simply the values you calculated and not the magnitudes.
Since $|T|<|V|$, $\quad E<0$, which is in general not uncommon and in this case is an indicator that the ISS is in a stable, bound orbit around the Earth, as can be seen in the included image.
For an non-relativistic object moving in a circular orbit - having maximum angular momentum that is - in a $1/r$ potential $V=-\frac{1}{2}K$. More generally the Virial theorem holds. Note that the total energy of the object is not zero if you add the rest energy $E=mc^2$.