# Find the work from Newton's law of gravitation

• amazofntheab
Earth) and a very low apogee (far from the Earth) is an example.In summary, Newton's Law of Gravitation states that two bodies with masses M and N attract each other with a force where r is the distance between the bodies and G is the gravitational constant.f

#### amazofntheab

Homework Statement
Newton's Law of Gravitation states that two bodies with masses M
and N
attract each other with a force where r
is the distance between the bodies and G
is the gravitational constant.

Use Newton's Law of Gravitation to compute the work (in J) required to launch a 1000-kg satellite vertically to an orbit 1000 km high. You may assume that Earth's mass is 5.98×1024kg
and is concentrated at its center. Take the radius of the Earth to be 6.37×106m
and G=6.67×10−11N⋅m^2/kg^2.
Relevant Equations
F = GMN/r^2, W=F*d
What I did was just sub in the numbers and convert km to m. So (6.67E-10)(1000)(5.98E24)/(1,000,000+6.37E6)^2 * (1,000,000+6.37E6) So it's just (6.67E-10)(1000)(5.98E24)/(1,000,000+6.37E6). I thought this was a straightforward problem, but it seems that it is not. What am I missing?

Homework Statement:: Newton's Law of Gravitation states that two bodies with masses M
and N
attract each other with a force where r
is the distance between the bodies and G
is the gravitational constant.

Use Newton's Law of Gravitation to compute the work (in J) required to launch a 1000-kg satellite vertically to an orbit 1000 km high. You may assume that Earth's mass is 5.98×1024kg
and is concentrated at its center. Take the radius of the Earth to be 6.37×106m
and G=6.67×10−11N⋅m^2/kg^2.
Relevant Equations:: F = GMN/r^2, W=F*d

What I did was just sub in the numbers and convert km to m. So (6.67E-10)(1000)(5.98E24)/(1,000,000+6.37E6)^2 * (1,000,000+6.37E6) So it's just (6.67E-10)(1000)(5.98E24)/(1,000,000+6.37E6). I thought this was a straightforward problem, but it seems that it is not. What am I missing?
Sub in the numbers to what equation?

Homework Statement:: Newton's Law of Gravitation states that two bodies with masses M
and N
attract each other with a force where r
is the distance between the bodies and G
is the gravitational constant.

Use Newton's Law of Gravitation to compute the work (in J) required to launch a 1000-kg satellite vertically to an orbit 1000 km high. You may assume that Earth's mass is 5.98×1024kg
and is concentrated at its center. Take the radius of the Earth to be 6.37×106m
and G=6.67×10−11N⋅m^2/kg^2.
You don't make it easy for readers to understand the above. I assume that "5.98×1024kg" means 5.98 x ##10^{24}## kg and "6.67×10−11N⋅m^2/kg^2" means 6.67× ##10^{-1}## N⋅m^2/kg^2. At the very least, why didn't you include ^ symbols for the exponents?
amazofntheab said:
Relevant Equations:: F = GMN/r^2, W=F*d

What I did was just sub in the numbers and convert km to m. So (6.67E-10)(1000)(5.98E24)/(1,000,000+6.37E6)^2 * (1,000,000+6.37E6) So it's just (6.67E-10)(1000)(5.98E24)/(1,000,000+6.37E6). I thought this was a straightforward problem, but it seems that it is not. What am I missing?
Surely you don't really mean "1,000,000+6.37E6" as you've written three times above.

What am I missing?
You are missing the fact that ##W=Fd## is the work done on an object by a constant force. Here the force is not constant but depends on the distance from the Earth's center. It might be useful to note that the work done by gravity is the negative of the change in potential energy. I suggest that you use the work-energy theorem.

• Kumail Haider, MatinSAR, FactChecker and 1 other person
##W = \int F(r) \, \mathrm{d}r##

• scottdave, FactChecker and jbriggs444
##W = \int F(r) \, \mathrm{d}r##
I think there is more to the problem than just doing the integral or, alternatively, figuring out the change in gravitational potential energy. The problem is asking "to compute the work (in J) required to launch a 1000-kg satellite vertically to an orbit 1000 km high." I am no rocket scientist, but I do know that gravity does not do work that puts satellites in orbit.

Presumably, the satellite is placed in orbit and does not fall back to Earth. This means that its kinetic energy relative to the center of the Earth has changed to that appropriate at that orbit. That kinetic energy can only come from the thrust which is unknowable. We can set the initial kinetic (relative to the Earth's center) equal to zero if we assume a launch from one of the poles and calculate the orbital kinetic energy assuming a circular orbit. At that point we can apply the work-energy theorem, $$W_g+W_{\text{thrust}}=\frac{1}{2}mv_{\text{orb}}^2$$ and solve for ##W_{\text{thrust}}## which is what I would consider as the work required to launch the satellite to that orbit.

• MatinSAR, FactChecker and malawi_glenn
Presumably, the satellite is placed in orbit and does not fall back to Earth.
It's unclear. It says "vertically", which would not result in entering orbit, and "to an orbit", as though orbit is simply a place, not "into orbit".

• scottdave, MatinSAR and FactChecker
It's unclear. It says "vertically", which would not result in entering orbit, and "to an orbit", as though orbit is simply a place, not "into orbit".
Right, I thought about that interpretation because of the word "vertical". But then why use "orbit" instead of "maximum altitude"? What kind of launch is this in which the satellite is expected to fall back to the Earth?
Ans: A launch that wastes the taxpayers' money.

Whatever the author had in mind, I think my interpretation makes more sense and a better problem.

• MatinSAR
Right, I thought about that interpretation because of the word "vertical". But then why use "orbit" instead of "maximum altitude"? What kind of launch is this in which the satellite is expected to fall back to the Earth?
Ans: A launch that wastes the taxpayers' money.

Whatever the author had in mind, I think my interpretation makes more sense and a better problem.
Not all orbits are circular. An elliptical orbit with a very large eccentricity would be consistent with a [near-]vertical launch. Since all of the Earth's mass is considered to lie at its center, the fact that such an orbit would ordinarily intersect with the crust of the planet does not pose a problem.

• FactChecker
Since all of the Earth's mass is considered to lie at its center, the fact that such an orbit would ordinarily intersect with the crust of the planet does not pose a problem.
OK, I understand now. We consider all of the Earth's mass concentrated at the center, yet the launch occurs 6400 km away from the center where there is no mass to support it.

• jbriggs444
OK, I understand now. We consider all of the Earth's mass concentrated at the center, yet the launch occurs 6400 km away from the center where there is no mass to support it.
It is a Wile E Coyote launch platform. If he does not look down, it will work just fine.

Saves money on blast deflectors too.

• • MatinSAR and kuruman
I think there is more to the problem than just doing the integral or, alternatively, figuring out the change in gravitational potential energy. The problem is asking "to compute the work (in J) required to launch a 1000-kg satellite vertically to an orbit 1000 km high." I am no rocket scientist, but I do know that gravity does not do work that puts satellites in orbit.

Presumably, the satellite is placed in orbit and does not fall back to Earth. This means that its kinetic energy relative to the center of the Earth has changed to that appropriate at that orbit. That kinetic energy can only come from the thrust which is unknowable. We can set the initial kinetic (relative to the Earth's center) equal to zero if we assume a launch from one of the poles and calculate the orbital kinetic energy assuming a circular orbit. At that point we can apply the work-energy theorem, $$W_g+W_{\text{thrust}}=\frac{1}{2}mv_{\text{orb}}^2$$ and solve for ##W_{\text{thrust}}## which is what I would consider as the work required to launch the satellite to that orbit.

I interpret "Use Newton's Law of Gravitation to compute the work (in J) required to launch a 1000-kg satellite vertically to an orbit 1000 km high." that it is only the vertical work that is asked for. It does not read placed in orbit. I think we should imagine that there is an orbit 1000 km above the surface of the Earth and calculate the amount of work needed to get to that orbit.

The "work required" is ambiguous without saying what force is doing the work on what system. We can safely assume that the system is the satellite, but what about the force that is doing the work on it? You say
that it is only the vertical work that is asked for
but you also omit specifying the force that does this "vertical work." There are two candidates, gravity and engine thrust.

Is it gravity? I don't think so because we don't use the force of gravity to do the "required" work that results in the necessary kinetic energy to launch satellites. Therefore it must be thrust and that's where we run into problems. The satellite's state of motion when it gets up there is important. As per the work-energy theorem, the work done by the thrust depends on the kinetic energy of the satellite at 1000 km.

Even if one interprets "orbit" as "altitude", the problem does not specify that the satellite reaches 1000 km with zero kinetic energy. That is an additional assumption that needs to be made. I preferred the assumption that the satellite is launched so that it goes into a circular orbit at 1000 km.

I am with you that it may very well be that the author expected the solver to do a line integral of the gravitational force. If so, why didn't the author just say so and why is the vertical launch stipulation necessary? When the satellite reaches 1000 km, the work done by gravity is independent of the launch direction. If the intent of this problem is to get one to do the line integral, that intent is obfuscated by all this ambiguous orbit and satellite stuff. I think that the author did not think this through completely.

To summarize, the formulation of this problem is unclear because
1. It uses the word "orbit" in a way that can be confused with "altitude."
2. It does not specify what force is doing the "required" work.
3. It does not specify directly or indirectly the satellite's state of motion at 1000 km.

I give this problem a D.

• nasu
Right, I thought about that interpretation because of the word "vertical". But then why use "orbit" instead of "maximum altitude"? What kind of launch is this in which the satellite is expected to fall back to the Earth?
Ans: A launch that wastes the taxpayers' money.

Whatever the author had in mind, I think my interpretation makes more sense and a better problem.
Maybe it doesnt waste money. Perhaps it is intended to take some pictures or gather other data from that altitude, then return to earth.

If one is measuring the capacity of a self-contained rocket motor in Joules and has an Earth-like planet arranged as specified (all mass concentrated in the center) then it is amusing to note that the rocket energy required to reach an altitude of 1000 km above the "surface" is approximately zero.

One simply arranges for a near-polar launch site, allows the craft to free fall to achieve an appropriately large velocity and then performs a tiny burn. The Oberth effect does the rest.

[You want to avoid an equatorial launch because that squanders delta V. You want to avoid a perfectly polar launch because you do not want to get a 2 centimeter hole punched in your rocket by the central black hole where the mass of the Earth went]

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