Answer:
v = √{2(GmE(1/RE - 1/(RE + h)]}
Explanation:
From the law of conservation of energy, the total kinetic + potential energy at h = total kinetic + potential energy at the surface of the earth(h = 0)
So, K + U = K' + U'
where K = kinetic energy of hammer at h = 0
U = gravitational potential energy of hammer at h = -GmEm/(RE + h)
K' = kinetic energy of hammer at the earth's surface (h = 0) = 1/2mv²
U = gravitational potential energy of hammer at earth's surface (h = 0) = -GmEm/RE
So, K + U = K' + U'
0 + [-GmEm/(RE + h)] = 1/2mv² + [-GmEm/RE]
0 - GmEm/(RE + h) = 1/2mv² - GmEm/RE
collecting like terms,we have
1/2mv² = GmEm/RE - GmEm/(RE + h)
Factorizing GmE and multiplying both sides by 2, we have
v² = 2(GmE(1/RE - 1/(RE + h)]
taking square-root of both sides, we have
v = √{2(GmE(1/RE - 1/(RE + h)]}
The expression for the speed of the hammer will be:
[tex]V=\sqrt{2(Gm_e(\dfrac{1}{R_e}-\dfrac{1}{R_e+h}) }[/tex]
The law of conservation says that the total energy at the height h and the total energy at the surface of the earth will remain constant.
From the law of conservation of energy
[tex]KE+PE=KE'+PE'[/tex]
where
KE = kinetic energy of hammer at h = 0
PE = gravitational potential energy of hammer at h height
[tex]PE=\dfrac{-Gm_em}{R_e+h}[/tex]
KE' = kinetic energy of hammer at the earth's surface (h = 0)
[tex]KE'=\dfrac{1}{2} mv^2[/tex]
PE' = gravitational potential energy of hammer at earth's surface (h = 0)
[tex]PE'=\dfrac{Gm_em}{R_e}[/tex]
Now by putting the values
[tex]KE+PE=KE'+PE'[/tex]
[tex]0+ \dfrac{-Gm_em}{R_e+h}[/tex] [tex]=\dfrac{1}{2} mv^2[/tex] [tex]-\dfrac{Gm_em}{R_e}[/tex]
[tex]\dfrac{1}{2} mv^2=\dfrac{Gm_em}{R_e} -\dfrac{Gm_em}{R_e+h}[/tex]
[tex]v^2=2Gm_e(\dfrac{1}{R_e} -\dfrac{1}{R_e+h} )[/tex]
[tex]v=\sqrt{2Gm_e(\dfrac{1}{R_e} -\dfrac{1}{R_e+h} )}[/tex]
Thus the expression for the speed of the hammer will be:
[tex]V=\sqrt{2(Gm_e(\dfrac{1}{R_e}-\dfrac{1}{R_e+h}) }[/tex]
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