How High Can an Object Reach if Projected Upward at 3.7 km/s?

In summary, using the energy conservation equation, the maximum height reached by an object projected upward from the surface of the Earth with an initial speed of 3.7 km/s can be calculated by substituting the values for the radius of the Earth, mass of the Earth, and gravitational constant into the formula. The resulting value is then added to the radius of the Earth to get the final height. In this case, the calculated maximum height is approximately 787.135 km.
  • #1
maniacp08
115
0
Gravity -- maximum height

An object is projected upward from the surface of the Earth with an initial speed of 3.7 km/s. Find the maximum height it reaches.
m

Ive used the Energy conservation
Kf + Uf = Ki + Ui
1/2 mvf^2 - G*M_e*m/rf = 1/2 mvi^2 - G*M_e*m/ri
= 0 - G*M_e*m/rf = 1/2 mvi^2 - G*M_e*m/R_e

R_e = radius of Earth = 6.37 x 10^6m
M_e = mass of Earth = 5.98 x 10^24kg
G = 6.67 x 10^-11 N*m^2/kg^2

1/radius_f = -Vi^2/2*G*M_e + 1/R_e
1/radius_f = (-3700 m/s)^2 / 2(6.67 x 10^-11)(5.98 x 10^24) + 1/6.37 x 10^6
=13690000 / 7.97732 x 10^13 + 1/6.37 x 10^6
1.716115187^-7 + 1.569858713^-7 = 3.2859739^-7
1/3.2859739^-7 = 3043237.806
then I do
3043237.806 - 6.37 x 10^6 but that gives me a negative height.

Can someone help me? I've been going at this problem for hours.
 
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  • #2


Sorry, but this is really hard to follow. I find it much easier to write out the whole problem algebraically before subbing in any numbers, and also using consistant significant figures.

If you could do this first it would help a lot.
 
  • #3


This is the final equation:
1/radius_f = -Vi^2/2*G*M_e + 1/R_e

then I solve for radius_f then I divide it by 1

then I subtract it from the Earth's radius to get the height but it gives me a negative height.
 
  • #4


I'm not entirely sure why that is, but it could have something to do with order of operations (don't quote me on that). Using your formula, I would take a reciprocal of it first: [tex]R + r = \frac{2GM_e}{-v^2} + R [/tex]

From this we can see that the R terms cancel and we're left with r that we want to find. I have not solved this mathematically yet though.
 
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  • #5


well the textbook has the same problem explained but different velocity and it uses that equations and same steps I posted. But I don't know what I did wrong mathematically to give me a negative height. I was wondering if someone can work it out and see if they get the same answer or diff.
 
  • #6


maniacp08 said:
1/radius_f = -Vi^2/2*G*M_e + 1/R_e
1/radius_f = (-3700 m/s)^2 / 2(6.67 x 10^-11)(5.98 x 10^24) + 1/6.37 x 10^6
=13690000 / 7.97732 x 10^13 + 1/6.37 x 10^6

you go wrong right here: you forgot a minus sign in the first term on the rhs going from the 1st to 2nd line
 
  • #7


maniacp08 said:
An object is projected upward from the surface of the Earth with an initial speed of 3.7 km/s. Find the maximum height it reaches.
m

Ive used the Energy conservation
Kf + Uf = Ki + Ui
1/2 mvf^2 - G*M_e*m/rf = 1/2 mvi^2 - G*M_e*m/ri
= 0 - G*M_e*m/rf = 1/2 mvi^2 - G*M_e*m/R_e

R_e = radius of Earth = 6.37 x 10^6m
M_e = mass of Earth = 5.98 x 10^24kg
G = 6.67 x 10^-11 N*m^2/kg^2

1/radius_f = -Vi^2/2*G*M_e + 1/R_e
1/radius_f = (-3700 m/s)^2 / 2(6.67 x 10^-11)(5.98 x 10^24) + 1/6.37 x 10^6
=13690000 / 7.97732 x 10^13 + 1/6.37 x 10^6
1.716115187^-7 + 1.569858713^-7 = 3.2859739^-7
1/3.2859739^-7 = 3043237.806
then I do
3043237.806 - 6.37 x 10^6 but that gives me a negative height.

Can someone help me? I've been going at this problem for hours.

Gravitational Potential is given by -GMm/r
So let u = GM so that

1/2mV2 - um/R = - um/(R + h)

V2/(2*u) = - 1/(R + h) + 1/R = h/(R2 +R*h)

You might recognize then in the near field with R >> h :

V2 = 2 * (u/(R2 +R*h)) * h ≈ 2*(g)*h

Rearranging:
2*u*h = V2R2 + V2*R*h

Solving for h then I get
h = V2R2/(2*u - V2*R)

Using km and u in km as ≈ 4*105 and R as 6400 km V as 3.7 km/s then ...
 
  • #8


h = V^2R^2/(2*u - V^2*R)

h = 13.69 * 40960000 / (800000 - 87616)
h = 560742400 / 712384
h = 787.1350283 km
h = .7871350283 m

Do I add this to the Radius of the earth?
 
  • #9


maniacp08 said:
h = V^2R^2/(2*u - V^2*R)

h = 13.69 * 40960000 / (800000 - 87616)
h = 560742400 / 712384
h = 787.1350283 km
h = .7871350283 m

Do I add this to the Radius of the earth?

The way it was originally constructed h was height above the radius.
1/2mV2 - um/R = - um/(R + h)

Better check your conversion however. The bolded line looks OK. The one below - not.

And significant digits wise you should round likely to the nearest km.

The value of u for Earth for instance I already rounded.

See: http://en.wikipedia.org/wiki/Gravitational_constant#The_GM_product
 
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1. What is the maximum height that an object can reach when thrown upwards?

The maximum height that an object can reach when thrown upwards is determined by the initial velocity and the force of gravity. It can be calculated using the formula h = (v2sin2θ)/2g, where h is the maximum height, v is the initial velocity, θ is the angle of projection, and g is the acceleration due to gravity.

2. How does air resistance affect the maximum height of an object?

Air resistance, or drag, can decrease the maximum height of an object by opposing its motion. This force increases as the object moves faster and can ultimately prevent it from reaching its maximum height. However, for small and dense objects like baseballs, air resistance is negligible and does not significantly impact the maximum height.

3. Can an object reach an infinite height when thrown upwards?

No, an object cannot reach an infinite height when thrown upwards. This is because as the object moves higher, the force of gravity decreases, and eventually, it will reach a point where the force of gravity equals the object's weight, causing it to stop and fall back to the ground.

4. Does the maximum height of an object change with the location on Earth?

Yes, the maximum height of an object does change with the location on Earth. This is because the acceleration due to gravity varies with the distance from the center of the Earth. Objects will reach a greater height at the equator compared to the poles due to the Earth's rotation and shape.

5. How does the mass of an object affect its maximum height?

The mass of an object does not affect its maximum height when thrown upwards. This is because the force of gravity acts on all objects equally, regardless of their mass. However, a lighter object may appear to reach a greater height due to its faster velocity, as determined by the equation h = (v2sin2θ)/2g.

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