Question

A satellite moves in a circular orbit around
the Earth at a speed of 6.3 km/s.
Determine the satellite’s altitude above
the surface of the Earth. Assume the
Earth is a homogeneous sphere of radius
6370 km and mass 5.98 × 1024 kg. The
value of the universal gravitational constant
is 6.67259 × 10−11 N · m2/kg2.
Answer in units of km

Answer 1

Well it is a physics problem but I tried using the gravitational constant one I cant figure it out

Answer 2

I know its a physics problem, but what topic?

Answer 3

I'm not sure honestly :/

Answer 4

Strange, where did you pick this from then?

Answer 5

It's from a review for my physics test tomorrow

Answer 6

Sure that's all the information?

Answer 7

That's all it gave me I've been trying to figure out how to get the mass of the satellite but I haven't figure it out since they once gave the speed

Answer 8

Give me the link to the place you got this question from.

Answer 9

Hmmm idt I can because it's through the university and you need a password to get in

Answer 10

You need the formula of orbital velocity,
\[v=\sqrt{\frac{GM}{R}}\] with M the mass of the planet and R the radius of the orbit.

Answer 11

The radius of the orbit, also, is,
\[R=R_T+h\] where R_T is the radius of the planet and h is the altitude above the surface of the planet.

Answer 12

Can you continue with this information?

Answer 13

You should solve for h, and obtain something like
\[h=\frac{GM}{v^2}-R_T\]

Answer 14

Ok thanks I'll try that :)

Answer 15

;)

Answer 16

You can deduce that you should use the orbital velocity because it is a circular movement, so, the gravitational force is a centripetal force, in modulus,
\[G\frac{mM}{R^2}=m\frac{v^2}{R}\]Solving for v, you obtain the orbital velocity.

Answer 17

but m would be the mass of the satellite?

Answer 18

Exact. The mass of the satellite as you can see can be eliminated, and doesn't matter.

Answer 19

\[G\frac{mM}{R^2}=m\frac{v^2}{R}\Rightarrow G\frac{M}{R^2}=\frac{v^2}{R}\]

Answer 20

I got 250281 for the velocity so I use this one or the 6.3 km/s for the previous problem?

Answer 21

No, the velocity cannot change, it is v=6.3 Km/s

Answer 22

Use the equation I gave you,
\[h=\frac{GM}{v^2}-R_t=\frac{6.67\cdot10^{-11}\cdot5.98\cdot10^{24}}{(6.3\cdot10^3)^2}-6.37\cdot10^6\]

Answer 23

@John_ES Escape velocity is \[V_{Escape} = \sqrt{\frac{ 2GM }{ R }}\]

Answer 24

You don't need escape velocity, you need orbital velocity because you the satellite doesn't need to go out the Earth. The satellite is orbiting the Earth so it has orbital velocity.

Answer 25

I haven't studied up to that point.

Answer 26

You can check a brief introduction here,
http://en.wikipedia.org/wiki/Orbital_speed

Answer 27

The solution I obtain is
\[h=10^{13}\ m\]
Remember to put all distances in meters.

Answer 28

Hmmm I got 3.67953*10^6

Answer 29

You're right I don't put the right numbers. ;)

Answer 30

In my calculator XD.