What is the objects velocity when it hits the earth

In summary, the conversation discusses a problem involving an object released in space at a distance of 3(6,37*10^6) from the center of the earth, without any initial velocity. The first question is about the object's velocity when it hits the earth, which has already been solved. The next question is split into two parts, asking about the time it takes for the object to reach the halfway point towards the earth and the total time it takes to hit the earth. The conversation then moves on to discussing the equation for conservation of energy and how to incorporate speed into it. The person asking for help has come up with an equation involving potential energy, but is unsure if it is correct. The expert suggests using the equation
  • #1
martine80
13
0
Hi, I am a student from norway, and I am strugeling with this problem, hope someone can help me og guide me in the direction:

A object is brought out in space in a distance of 3(6,37*10^6) from the senter of the earth. not any velocity when the object is released in space.
1) What is the objects velocity when it hits the Earth ( this I've solved)
2)
i)how long does the object use halfway towards the earth?
ii)how log does it take until the object hits the earth?

HELP PLEASE!
 
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  • #2
Write down the equation for conservation of energy, with speed written as a derivative.
 
  • #3
George Jones said:
Write down the equation for conservation of energy, with speed written as a derivative.
how do i do that?
so far I've come up with this:

2*R(radius of the earth)=3R-integrate from 0 to t(1/2(integrate from 3R to 2R(-G*M(Earth)/r^2)) *t^2
 
  • #4
martine80 said:
how do i do that?
so far I've come up with this:

2*R(radius of the earth)=3R-integrate from 0 to t(1/2(integrate from 3R to 2R(-G*M(Earth)/r^2)) *t^2

I'm not sure what you're doing here.

To make sure things are OK, just write down the conservation of energy equation, and leave the integrals until later.
 
  • #5
I use : 1/2 mv^2
and Gravital Potential energy, and unerversial of gravity = Gm(1)m(2)/r^2
 
  • #6
martine80 said:
I use : 1/2 mv^2
and Gravital Potential energy, and unerversial of gravity = Gm(1)m(2)/r^2

Let R be the radius of the Earth, and suppose that the object is a distance r from the centre of the Earth. If the object starts at rest from 3R, what is the *equation* for conservation of energy.
 
  • #7
George Jones said:
Let R be the radius of the Earth, and suppose that the object is a distance r from the centre of the Earth. If the object starts at rest from 3R, what is the *equation* for conservation of energy.

:cry: I don't know, I've sat down with a teacher and he didnt know eather , just the equation i told you about
 
  • #8
martine80 said:
:cry: I don't know, I've sat down with a teacher and he didnt know eather

?

Let K be kinetic energy and V be gravitational potential energy.

K_initial + V_intitial = K_final + V_final

Can you substitute expressions for each of the terms in this equation?
 
  • #9
George Jones said:
?

Let K be kinetic energy and V be gravitational potential energy.

K_initial + V_intitial = K_final + V_final

Can you substitute expressions for each of the terms in this equation?

I know, but i toght i was suppose to use potentional energi in this ? I am not sure... I am about to give up now. As far as i know I am supose to end up with an integral I have to use Maple to solve ...
 
  • #10
martine80 said:
I know, but i toght i was suppose to use potentional energi in this ?

:confused: Again, I'm confused. My equation involves potential energy.

As far as i know I am supose to end up with an integral I have to use Maple to solve ...

If you'll substitute expressions (that I think you know) into the equation that I wrote previously, I'll help you get the integral, but, since this is homework (and thus should be in the homework section), I'm, at this point, not going to write down the answer.
 
  • #11
hehe, no its not homework, I don't have to do this ... Its voluntary

I come jp with this :

1/2mv(0)^2+ G M(earth) m(object) (1/r1-1/r2)=1/2mv^2+ G M(earth) m(object) (1/r1-1/r2)
Im sure its wrong :(
 
  • #12
martine80 said:
hehe, no its not homework, I don't have to do this ... Its voluntary

I come jp with this :

1/2mv(0)^2+ G M(earth) m(object) (1/r1-1/r2)=1/2mv^2+ G M(earth) m(object) (1/r1-1/r2)
Im sure its wrong :(

Almost.

If the object starts from rest, then, initially, v = 0, so K_initial = 0. This gets rid of one of the terms in the equation.

Also, the gravitation potential energy (of the system) when the object is a distance r from Earth's centre is

V = - G M m/r^2.

Use r_initial and r_final to get V_initial and V_final.

Because of work and family commitments, I might not be able to get back to this thread until tomorrow.
 
  • #13
George Jones said:
Almost.

If the object starts from rest, then, initially, v = 0, so K_initial = 0. This gets rid of one of the terms in the equation.

Also, the gravitation potential energy (of the system) when the object is a distance r from Earth's centre is

V = - G M m/r^2.

Use r_initial and r_final to get V_initial and V_final.

Because of work and family commitments, I might not be able to get back to this thread until tomorrow.

V initial = 0 and Vfinal is 2GM(E)/3radius of the Earth ...
Thank u for yore time , Ill let you know by tomorrow if I've solved the problem!
 

1. What is the definition of velocity?

Velocity is a measure of the rate of change of an object's position over time. It is a vector quantity, meaning it has both magnitude (speed) and direction.

2. How is velocity different from speed?

While both velocity and speed measure how fast an object is moving, velocity also includes the direction of motion. Speed is a scalar quantity, meaning it only has magnitude and not direction.

3. How is the velocity of an object calculated?

The velocity of an object is calculated by dividing the change in position (displacement) by the change in time. It can also be calculated by multiplying the object's speed by the direction in which it is moving.

4. Does an object's velocity change as it falls towards the earth?

Yes, an object's velocity changes as it falls towards the earth due to the effect of gravity. As the object falls, its speed increases and its direction changes towards the center of the earth.

5. What factors can affect an object's velocity when it hits the earth?

The factors that can affect an object's velocity when it hits the earth include the object's initial velocity, air resistance, and the force of gravity acting on the object. The mass and shape of the object can also play a role in its final velocity upon impact.

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