What is the Weight of a Spacecraft at Various Distances from Earth?

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In summary: I figured it out.In summary, the weight of a spacecraft at a distance of 1.31E4 km from Earth is 1840490798 Newtons.
  • #1
skins266
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Homework Statement



The radius of the Earth is about 6.38E3 km. A 7.38E3 N spacecraft travels away from Earth. What is the weight of the spacecraft at the following disatnaces.

a)6.38E3 km
b)1.31E4 km

Homework Equations



F=GMeMship/r^2


The Attempt at a Solution



but that won't work because the mass of the ship is in N so I divide it by 9.8 and then the two radiuses need to be added so it will look like this

F = (6.67E-11)(5.98E24)(753.1)/(6.38E3+6.38E3)^2

=3E17/1.63E8

=1840490798

but that is not the right answer when I punch it into Webassign, and I figure once I get the first one the second one is exactly the same way. So what am I doing wrong?

Oh and the answers are in Newtons
 
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  • #2
For one thing, you'd better put the distances in meters. The same as the units of the other constants in that equation. You can also solve the problem more directly by considering ratios. E.g. for part a) compare the force at R with the force at 2R. You don't need any numbers to do that.
 
Last edited:
  • #3
All you have to do is to compute the force. The units of force are Newtons, and "weight" is defined as the gravitational force.

You're chief problem here is one of units. Here is a very good practice you should try to get in the habit of using: Always express the units in an equation. For example, don't just use [itex]6.67*10^{-11}[/itex], use the value with its units, [itex]6.67*10^{-11}\mathrm{m}^3/\mathrm{kg}/\mathrm{s}^2[/itex]. The same goes for the masses and distances. You have to make sure the units match.
 
  • #4
so, just in units, (m^3/kg/s^2)(kg)(?)/(km^2). So how do I convert m^3 to km? and what is the missing unit. Other than that I have the equation right?
 
  • #5
The equation is fine. km=1000*m. So km^3=10^9*m^3. But it's easier just to convert you km radii to m. Easier still to work out the answer using proportions. If the radius changes by a factor of K, by what factor does the force change?
 
  • #6
it would be by K
 
  • #7
skins266 said:
it would be by K

Nooo. What power is r in your equation? If R -> 2R then...
 
  • #8
then it would be multiplied by two and squared
 
  • #9
If you mean the force is divided by 4, yes. 1/R^2 -> 1/(2R)^2=(1/R^2)/4.
 
  • #10
but which one is R, the radius of the Earth added to the radius where the spaceship is at, or what?
 
  • #11
Think about it. a) says compare the force at radius R (radius of the earth), with the force at radius 2R.
 
  • #12
so I just multiply my answer by the number in a)?
 
  • #13
skins266 said:
so I just multiply my answer by the number in a)?

I don't know what that means. Gravity is an inverse square law. Doubling distance reduces the force by a factor of four. Etc.
 
  • #14
i didn't know that, but where I am going off track here. Since my equation was right shouldn't I have the right answer? Or is it the units thing that I can't figure out?
 
  • #15
It's the units thing. Your equation is correct. Just convert the km to m. 6.83E3km=6.83E6m.
 
  • #16
Never mind.
 

1. What is the "Tricky Spacecraft Problem"?

The "Tricky Spacecraft Problem" is a hypothetical scenario in which a spacecraft is approaching a planet on a trajectory that will cause it to crash. The goal is to find a solution that will allow the spacecraft to safely enter orbit around the planet.

2. Why is this problem considered tricky?

This problem is considered tricky because there are many variables and constraints that must be taken into account in order to find a successful solution. The spacecraft's speed, trajectory, and fuel supply all play a role in determining the best course of action.

3. What are some potential solutions to the "Tricky Spacecraft Problem"?

Some potential solutions include using the spacecraft's thrusters to adjust its speed and trajectory, utilizing the planet's gravity to alter the spacecraft's path, or deploying a parachute to slow down the spacecraft's descent.

4. How do scientists approach this problem?

Scientists approach this problem by using mathematical models and simulations to analyze different scenarios and determine the best course of action. They also take into account any limitations or constraints that may affect the spacecraft's capabilities.

5. Has the "Tricky Spacecraft Problem" been solved in real life?

While this specific problem has not been encountered in real life, similar situations have been successfully navigated by spacecraft engineers and scientists. NASA's Apollo 13 mission, for example, faced a critical situation when an explosion occurred on board the spacecraft, and engineers had to come up with a solution to safely return the astronauts to Earth.

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