# Calculate Mass of Satellite: Step-by-Step Guide

• songoku
In summary, the equation for orbital speed is given by m(v^2/r) = G(m1m2/r^2) and can be rearranged to find the orbital velocity as v = sqrt(GM/r). The equation for escape speed is given by KE1 + GPE1 = KE2 + GPE2 and can be used to find the escape velocity. However, in this particular problem, the mass of the satellite is not needed to find the escape velocity as it only depends on the distance from the center of the Earth. Changing the mass of the Earth or gravity would not affect the ratio of orbital energy to escape energy or the ratio of orbital velocity to escape velocity. Therefore, the given problem lacks necessary information to find
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

songoku said:
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
haruspex said:
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

songoku said:
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.

SammyS and songoku
Thank you very much haruspex

songoku said:
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.

songoku
jbriggs444 said:
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##

valenumr said:
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.

jbriggs444 said:
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.

valenumr said:
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.

jbriggs444 said:
@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 said:
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.

jbriggs444 said:
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

valenumr said:
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
jbriggs444 said:
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.

jbriggs444
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

mpresic3 said:
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.

## What is the formula for calculating the mass of a satellite?

The formula for calculating the mass of a satellite is:
Mass = (Gravitational Constant x Orbit Radius^2) / (Gravitational Acceleration x Period^2)

## What is the gravitational constant?

The gravitational constant, denoted by G, is a fundamental physical constant that represents the strength of the gravitational force between two objects. It is approximately equal to 6.674 x 10^-11 Nm^2/kg^2.

## What is the orbit radius of a satellite?

The orbit radius of a satellite is the distance between the center of the satellite and the center of the body it is orbiting. This can be measured in meters (m) or kilometers (km).

## What is the gravitational acceleration?

The gravitational acceleration, denoted by g, is the acceleration due to gravity at the surface of a celestial body. On Earth, it is approximately equal to 9.8 m/s^2.

## How do I calculate the period of a satellite?

The period of a satellite is the time it takes for the satellite to complete one full orbit around the body it is orbiting. It can be calculated using the formula:
Period = 2π x √(Orbit Radius^3 / Gravitational Acceleration)

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