What Formulas Should I Use to Calculate the Speed of an Object Before Impact?

[laker_gurl3]

help with physics! grav force

hello can someone help me on these questions?
i just need help with which formulas to use, i could not get any answer at all.

a stationary 25 kg object is released from a position 8.9x10^6m from the center of the earth. what's is the speed of the object before impact?

 

[Dr.Brain]

Apply Conservation Of Energy.Before and after the impact.

Initially Total Energy: - \frac {GMm}{x}

Final Energy : - \frac {GMm}{R} + \frac{1}{2}mv^2

Equate above and Solve.

BJ

 

[bumclouds]

I am a year 12 student so please bear with me - I could be horribly wrong.

PS: this is my first post on this forum!

You need to figure out the change in gravitational potential energy from the height you gave, to the surface of the earth.

You need to know that
the formula for for grav. potential energy is given by mgh
radius of Earth is 6.3*10^6m ( I think? )
gravitational field strength of Earth is given by GM/(R^2)


Lets figure out the G part of mgh at the greater height
[(6.67*10^-11)*25000] / ((8.9*10^6)^2)
that is.. G M / R ^2

equals 2.11*10^-20

subsitute that into MGH...

= mgh...
= 25000 * (2.11*10^-20) * (8.9*10^6)
gpe at greater height = 4.7 * 10^-9 Joules


the gravitational field strength at the Earth's surface is 9.8N/kg
so the GPE at surface of the Earth is

= M G H
= 25000 * (9.8*25) * (6.3*10^6)

equals ? something that doesn't look right?

this is converted to Kinetic Energy??... i think? oh god.. I am ready to get flamed badly

 

[Dr.Brain]

I am a year 12 student so please bear with me - I could be horribly wrong.

PS: this is my first post on this forum!

You need to figure out the change in gravitational potential energy from the height you gave, to the surface of the earth.

You need to know that
the formula for for grav. potential energy is given by mgh
radius of Earth is 6.3*10^6m ( I think? )
gravitational field strength of Earth is given by GM/(R^2)


Lets figure out the G part of mgh at the greater height
[(6.67*10^-11)*25000] / ((8.9*10^6)^2)
that is.. G M / R ^2

equals 2.11*10^-20

subsitute that into MGH...

= mgh...
= 25000 * (2.11*10^-20) * (8.9*10^6)
gpe at greater height = 4.7 * 10^-9 Joules


the gravitational field strength at the Earth's surface is 9.8N/kg
so the GPE at surface of the Earth is

= M G H
= 25000 * (9.8*25) * (6.3*10^6)

equals ? something that doesn't look right?

this is converted to Kinetic Energy??... i think? oh god.. I am ready to get flamed badly

mgh is only applicable for heights appreciably smaller than radius of the earth.

BJ

 

[HallsofIvy]

And please don't add a new question to someone else's thread- start your own thread.