Calculate Mass of Satellite: Step-by-Step Guide

[songoku]

Homework Statement

A satellite is orbiting earth. The satellite has orbital velocity of 5.9 km/s. If the minimum velocity for the satellite to escape earth is 14.6 km/s, what is the mass of the satellite if the satellite is located 3200 km above earth’s surface?

Relevant Equations

$F=m \frac{v^2}{r}$
$F=G\frac{m_1 . m_2}{r^2}$
$KE=\frac 1 2 mv^2$
$GPE=-G\frac{m_1.m_2}{r}$

I am not really sure what to do to find the mass of satellite.

Equation for orbital speed:
$$m \frac{v^2}{r}=G\frac{m_1 . m_2}{r^2}$$
$$v_{orbital}=\sqrt{\frac{GM}{r}}$$

Equation for escape speed:
$$KE_1+GPE_1=KE_2+GPE_2$$

I tried to take position 1 as the position where the satellite orbits and position 2 is at infinity (where both KE and GPE are zero) but I can't find the mass of the satellite.

How to approach this question? Thanks

 

[haruspex]

Homework Statement:: A satellite is orbiting earth. The satellite has orbital velocity of 5.9 km/s. If the minimum velocity for the satellite to escape Earth is 14.6 km/s, what is the mass of the satellite if the satellite is located 3200 km above earth’s surface?
Relevant Equations:: $F=m \frac{v^2}{r}$
$F=G\frac{m_1 . m_2}{r^2}$
$KE=\frac 1 2 mv^2$
$GPE=-G\frac{m_1.m_2}{r}$

I am not really sure what to do to find the mass of satellite.

Equation for orbital speed:
$$m \frac{v^2}{r}=G\frac{m_1 . m_2}{r^2}$$
$$v_{orbital}=\sqrt{\frac{GM}{r}}$$

Equation for escape speed:
$$KE_1+GPE_1=KE_2+GPE_2$$

I tried to take position 1 as the position where the satellite orbits and position 2 is at infinity (where both KE and GPE are zero) but I can't find the mass of the satellite.

How to approach this question? Thanks

A thought experiment: consider another such satellite next to the first. Is there any reason the second would not obey all the same conditions? Now consider them merged into a single satellite of twice the mass. Any reason that would not also satisfy all the same conditions?

 

[songoku]

A thought experiment: consider another such satellite next to the first. Is there any reason the second would not obey all the same conditions? Now consider them merged into a single satellite of twice the mass. Any reason that would not also satisfy all the same conditions?

No, so does it mean that there is no answer to this question since based on the equations the mass of satellite is irrelevant?

Thanks

 

[haruspex]

No, so does it mean that there is no answer to this question since based on the equations the mass of satellite is irrelevant?

Thanks

I'd say so.

 

[songoku]

Thank you very much haruspex

 

[jbriggs444]

Homework Statement:: A satellite is orbiting earth. The satellite has orbital velocity of 5.9 km/s. If the minimum velocity for the satellite to escape Earth is 14.6 km/s

It would appear that this is a problem that you made up yourself.

If 5.9 km/s yields a circular orbit then the required escape velocity at that altitude would be only $\sqrt{2}$ times as much: 8.3 km/s. The figure of 14.6 km/s is incorrect.

 

[valenumr]

It would appear that this is a problem that you made up yourself.

If 5.9 km/s yields a circular orbit then the required escape velocity at that altitude would be only $\sqrt{2}$ times as much: 8.3 km/s. The figure of 14.6 km/s is incorrect.

Does that work out for surface escape velocity? Also, thinking out of the box... No value is specified for $G$

 

[jbriggs444]

Does that work out for surface escape velocity?

Yes. Low Earth orbital velocity is 7.8 km/s. Low Earth escape velocity is $\sqrt{2}$ times that -- 11 km/s.

An object in a circular orbit always has half the kinetic energy required to escape.

 

[valenumr]

Yes. Low Earth orbital velocity is 7.8 km/s. Low Earth escape velocity is $\sqrt{2}$ times that -- 11 km/s.

An object in a circular orbit always has half the kinetic energy required to escape.

Sorry, I was joking a little. I think Earth escape is around 11 m/s, and solar system (from earth) is around 17 m/s. The question is sketchy unless you want to change gravity or the mass of the earth.

 

[jbriggs444]

Sorry, I was joking a little. I think Earth escape is around 11 m/s, and solar system (from earth) is around 17 m/s. The question is sketchy unless you want to change gravity or the mass of the earth.

@haruspex has identified the primary sketchiness well enough: None of the figures depend on the mass of the satellite.

Changing gravity or the mass of the Earth does nothing to address that concern.

 

[valenumr]

@haruspex has identified the primary sketchiness well enough. None of the figures depend on the mass of the satellite. Changing gravity or the mass of the Earth does nothing to address that concern.

Right. Also I meant km/s. The orbital velocity only depends on distance.

 

[valenumr]

Right. Also I meant km/s. The orbital velocity only depends on distance.

Well, okay, but wouldn't the orbital energy be higher (and therefore velocity vs distance) if Earth were more massive? Or of gravity were stronger. Sorry, I tend to talk fast and think slow.

 

[songoku]

It would appear that this is a problem that you made up yourself.

If 5.9 km/s yields a circular orbit then the required escape velocity at that altitude would be only $\sqrt{2}$ times as much: 8.3 km/s. The figure of 14.6 km/s is incorrect.

This is not the problem I made by myself. I also realized all the things you wrote, I even googled escape velocity of earth, which is 11.2 km/s. I also reverse calculated mass of Earth from given data and it did not match the actual mass of earth. I am just afraid I missed something, like misinterpreting the question, so I posted it here.

Thank you jbriggs444

 

[jbriggs444]

Well, okay, but wouldn't the orbital energy be higher (and therefore velocity vs distance) if Earth were more massive? Or of gravity were stronger. Sorry, I tend to talk fast and think slow.

Yes. Orbital velocity and energy would be higher. And escape velocity and energy would be higher.

Say that you quadruple the mass of the planet. Or quadruple gravity. But you still arrange for a circular orbit at the prescribed radius. Escape energy has quadrupled, obviously. So escape velocity has doubled. Now we need centrifugal force ($\frac{mv^2}{r}$) to quadruple. So orbital velocity has doubled. And orbital energy has quadrupled.

So the ratio of orbital energy to escape energy has remained constant. As has the ratio of orbital velocity to escape velocity.

 

[valenumr]

Yes. Orbital velocity and energy would be higher. And escape velocity and energy would be higher.

Say that you quadruple the mass of the planet. Or quadruple gravity. Escape energy has quadrupled, obviously. So escape velocity has doubled. Now we need centrifugal force ($\frac{mv^2}{r}$) to quadruple. So orbital velocity has doubled. And orbital energy has quadrupled.

So the ratio of orbital energy to escape energy has remained constant. As has the ratio of orbital velocity to escape velocity.

I was trying to make the numbers given make sense. But obviously it can't work.

 

[mpresic3]

In general, the motion in the gravity field can tell us nothing about the satellite mass. All masses fall with similar motion in the gravify field. I suggest this is a poorly phrased problem

 

[DaveC426913]

I suggest this is a poorly phrased problem

Just for fun and pedantry, I think it's only problematic because we are assuming the satellite is of negligible mass compared to Earth.

The math is beyond me but I wonder if there is as solution for that orbital v and escape v if the satellite's mass is a sizable fraction of the Earth.