Find the speed of a satellite at a distance R from Earth

In summary, the problem statement is asking for the velocity of a satellite at a distance of two from the center of the Earth. However, the equation that is given does not correctly express the centripetal force required for a circular trajectory. The answer is found by solving for v.
  • #1
Helly123
581
20

Homework Statement


Gravitational force exerted on mass m is GMm/r^2

2. Relevant equations
Orbital velocity at distance R from Earth = ##\sqrt gR##
Escape velocity = ##\sqrt 2gR##
gR = GM/R
Fc = mV^2/R
F =m.a

The Attempt at a Solution


1) express acceleration of gravity in terms of G, M , and R
F = m.a
a = F/m = GM/r^2

2) velocity of satellite at distance R from Earth's surface
I thought v = ##\sqrt gR##
But it is wrong.
I thought it is mv^2/R = GMm/R

3) mechanical energy for satellite in terms of G, m, R and EP at distance r infinity = 0
So, left EK. But i have to find the v first

Can i get a clue?
 
Physics news on Phys.org
  • #2
Hi,
Whaat is the problem statement ? And what are considered known variables ?
 
  • #3
BvU said:
Hi,
Whaat is the problem statement ? And what are considered known variables ?
Hi Sir.
The problem statement only that above. Variable is m, R, force of gravitational.
 
  • #4
Helly123 said:

Homework Statement


Gravitational force exerted on mass m is GMm/r^2

2. Relevant equations
Orbital velocity at distance R from Earth = ##\sqrt gR##
Escape velocity = ##\sqrt 2gR##
gR = GM/R
Fc = mV^2/R
F =m.a

The Attempt at a Solution


1) express acceleration of gravity in terms of G, M , and R
F = m.a
a = F/m = GM/r^2

2) velocity of satellite at distance R from Earth's surface
I thought v = ##\sqrt gR##
But it is wrong.
I thought it is mv^2/R = GMm/R

3) mechanical energy for satellite in terms of G, m, R and EP at distance r infinity = 0
So, left EK. But i have to find the v first

Can i get a clue?
##\sqrt{2gR}## and ##\sqrt{gR}##
 
  • #5
Helly123 said:
Hi Sir.
The problem statement only that above. Variable is m, R, force of gravitational.

But a satellite can have any speed at a point that is at a distance R from the center of the earth. If the trajectory must be circular (as I suspect is the intention in this exercise), then that makes things very different.
 
  • Like
Likes Helly123
  • #6
BvU said:
But a satellite can have any speed at a point that is at a distance R from the center of the earth. If the trajectory must be circular (as I suspect is the intention in this exercise), then that makes things very different.
Orbital velocity is a circular motion right?
 
  • #7
Helly123 said:
Orbital velocity is a circular motion right?
Not all orbits are circular, no. They can be circular, elliptical, parabolic or hyperbolic.

However, many orbits are approximately circular. If the problem stated in the title of this thread is to be answered given values for only G, M, m and r then, like @BvU, I suspect that the person who posed the question expected you to assume a circular orbit.
Helly123 said:
I thought it is ##\frac{mv^2}{R} = \frac{GMm}{R}##
Equating the required centripetal force for a circular trajectory of radius R with the actual centripetal force provided by gravity. Almost correct. You should check both formulas [at least one is wrong]. Then solve for v.
 
  • Like
Likes Helly123
  • #8
jbriggs444 said:
Not all orbits are circular, no. They can be circular, elliptical, parabolic or hyperbolic.

However, many orbits are approximately circular. If the problem stated in the title of this thread is to be answered given values for only G, M, m and r then, like @BvU, I suspect that the person who posed the question expected you to assume a circular orbit.

Equating the required centripetal force for a circular trajectory of radius R with the actual centripetal force provided by gravity. Almost correct. You should check both formulas [at least one is wrong]. Then solve for v.
I missed it.
mv^2/R = GMm/R^2
V = ##\sqrt{GM/R}##
However, it is still wrong
 
  • #9
Helly123 said:
I missed it.
mv^2/R = GMm/R^2
V = ##\sqrt{GM/R}##
However, it is still wrong
The answer is ##\sqrt{gR/2}## for number 2. Number 2 including 3, it says to be expressed in terms of r, G
 
  • #10
You are leaving us with not much to go on. The original post does not contain the problem statement. No subsequent post does either.
 
  • #11
jbriggs444 said:
You are leaving us with not much to go on. The original post does not contain the problem statement. No subsequent post does either.
I am sorry.. but i have added it. That is what it says
 
  • #12
The problem statement might ask the speed of satellite at distance R from the surface of Earth (R is the radius of Earth).
 
  • Like
Likes Helly123
  • #13
ehild said:
The problem statement might ask the speed of satellite at distance R from the surface of Earth (R is the radius of Earth).
Which means the total R is 2R?
 
  • #14
Better to use two different symbols for two different variables. The phrase 'Which means the total R is 2R?' looks really weird to me (but I understand what you mean).

Why not work it out ?
 
  • Like
Likes Helly123
  • #15
So, r = 2R
r is distance from satellite to Earth's center
Btw, mechanical energy is Ek. The answer is -mGR/4
How can Ek be negative?
 
  • #16
When object has centripetal force mv^2/r,
The radius r is measured in regards to what? If satellite, is due to Earth's center or Earth's surface?
 
  • #17
Helly123 said:
When object has centripetal force mv^2/r,
The radius r is measured in regards to what? If satellite, is due to Earth's center or Earth's surface?
What is the centripetal force in general? When a particle moves along a circle of radius r?
 
  • #18
ehild said:
What is the centripetal force in general? When a particle moves along a circle of radius r?
Ok it is in regard to the center
 
  • #19
Helly123 said:
Ok it is in regard to the center
Yes. And where is the center of the orbit?
 
  • #20
ehild said:
Yes. And where is the center of the orbit?
Earth's center?
Why Ek can be negative? Do you know it?
 
  • #21
If you mean Ek kinetic energy, it can not be ó negative.
For a bound orbit, the total mechanical energy is negative, and the satellite does not reach infinity. What is the total mechanical energy on the orbit of radius r? On an orbit of radius 2R? What happens if r increases to infinity?
 
  • #22
ehild said:
If you mean Ek kinetic energy, it can not be ó negative.
For a bound orbit, the total mechanical energy is negative, and the satellite does not reach infinity. What is the total mechanical energy on the orbit of radius r? On an orbit of radius 2R? What happens if r increases to infinity?
Satellite does not reach infinity?
If r increases to infinity, gravitational force = 0, Fc = 0 ?
 
  • #23
The total mechanical energy does not change.
The problem states that the potential energy is zero at infinity. That means the satellite is not under the influence of other planets (Sun for example) even at far away from the Earth. What is the gravitational potential of the Earth at distance r ? What is the total mechanical energy Ep+Ek of the satellite along an orbit of radius r? The kinetic energy is 1/2 mv2. It can not be negative. The satellite can reach infinity if the speed exceeds the escape velocity. You calculated the speed on an orbit of radius 2R . How is it related to the escape velocity?
 
  • Like
Likes Helly123
  • #24
ehild said:
The total mechanical energy does not change.
The problem states that the potential energy is zero at infinity. That means the satellite is not under the influence of other planets (Sun for example) even at far away from the Earth. What is the gravitational potential of the Earth at distance r ? What is the total mechanical energy Ep+Ek of the satellite along an orbit of radius r? The kinetic energy is 1/2 mv2. It can not be negative. The satellite can reach infinity if the speed exceeds the escape velocity. You calculated the speed on an orbit of radius 2R . How is it related to the escape velocity?
Ok.
Escape velocity
1/2mv^2 = EP of Earth
1/2mv^2 = GMm/R
v = ##\sqrt{2GM/R}## = ##\sqrt{2gR}##
R = Earth's radius

Ep of Earth at distance r
= GMm/r = mgR^2/2R = mgR/2

Kinetic energy during orbitting is obtained by using escape velocity?
 
  • #25
Helly123 said:
Ok.
Escape velocity
1/2mv^2 = EP of earth
1/2mv^2 = GMm/R
v = ##\sqrt{2GM/R}## = ##\sqrt{2gR}##
R = Earth's radius

Ep of Earth at distance r
= GMm/r = mgR^2/2R = mgR/2

Kinetic energy during orbitting is obtained by using escape velocity?
No, it is calculated from the speed along the orbit, you have already.
What is the sign of the potential energy? The force points towards the center.
 
  • #26
ehild said:
No, it is calculated from the speed along the orbit, you have already.
What is the sign of the potential energy? The force points towards the center.
ehild said:
No, it is calculated from the speed along the orbit, you have already.
What is the sign of the potential energy? The force points towards the center.
is it negative? But why?
 
  • #27
Helly123 said:
is it negative? But why?
Where is the reference point where potential energy is zero?
 
  • #28
jbriggs444 said:
Where is the reference point where potential energy is zero?
In the center of the earth? Isn't it?
 
  • #29
Helly123 said:
In the center of the earth? Isn't it?
Using a reference point at the center of the earth, the potential energy of a satellite would be positive. The conventional reference point is infinitely far away instead.

You can read that from the formula for potential energy. If the potential energy is given by$$PE=-\frac{GMm}{r}$$then the potential energy is zero when r is infinite.

The choice of reference point is purely arbitrary. Only differences in potential energy are physically meaningful. Placing the reference point at infinity simplifies the above potential energy formula by eliminating the need to add a "+ C" at the end.
 
Last edited:
  • Like
Likes Helly123
  • #30
Helly123 said:
is it negative? But why?
Bodies released from rest, move in the direction so as their potential energy decrease.
It was said in the problem that the potential energy of the satellite is zero at infinity.
If it is in rest, the total mechanical energy is zero at infinity. Adding just a little push towards the Earth, it will move towards the Earth and its potential energy decreases. The potential energy decreases from zero, so it must be negative Ep=-GmM/r.
 
  • Like
Likes Helly123
  • #31
So, the total mechanical energy
= Ek + Ep = 1/2mv^2 + (-GmM/r) = gRm/4 - GmM/2R = -mgR/2. Is it?
 
  • #32
Helly123 said:
So, the total mechanical energy
= Ek + Ep = 1/2mv^2 + (-GmM/r) = gRm/4 - GmM/2R = -mgR/2. Is it?
No. Do not mix g and G. You know that GM/R2=g, so GM=gR2.
 
  • Like
Likes Helly123
  • #33
ehild said:
No. Do not mix g and G. You know that GM/R2=g, so GM=gR2.
Yes
 
  • #34
Helly123 said:
Yes
So what result did you get?
 
  • #35
Helly123 said:
So, r = 2R
r is distance from satellite to Earth's center
Btw, mechanical energy is Ek. The answer is -mGR/4
How can Ek be negative?

NO. The Mechanical Energy is the sum of the Gravitational Potential Energy plus the Kinetic Energy. The Kinetic Energy may be positive, but the Mechanical Energy value can still be negative, provided the Potential Energy is negative, and larger in size than the Kinetic Energy.
 
  • Like
Likes Helly123

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
16
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
855
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
184
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
Back
Top