Respuesta :
Answer:
a) U= -8.43 10 9 J b) 6.52 10⁹ J
Explanation:
a) The potential energy can be determined from the relationship between force and energy
F = - dU / dr
dU = - F dr
We replace and integrate. If we take the positive direction in the direction of dr, the force goes in the opposite direction (attraction) therefore it is negative
F = - G mM / r²
∫ U = G mM ∫ dr / r²
U = G mM (- 1 / r)
We evaluate
[tex]U_{f}[/tex] –U₀ = - G mM (1 /[tex]r_{f}[/tex] - 1 / r₀)
The reference system location is arbitrary, the most common choice is U₀ = 0 for r₀ = ∞
U = - G m M 1 / r
Let's reduce the magnitudes to the SI system
T = 702 min (60s / 1min) = 42120
R = 20100 km (1000m / 1km) = 2.01 10⁷ m
Let's calculate
U = -6.67 10⁻¹¹ 425 5.98 10²⁴ 1 / 2.01 10⁷
U= -8.43 10 9 J
b) The energy to place the satellite in orbit, This is equal to the mechanical energy of the system
Em = K + U
Em = ½ m v² +U
Since the satellite is in a circular orbit, the velocity module is constant
v = d / t
The distance of an orbit is the length of the circle and this time is called a period
d = 2π r
v = 2π r / T
We replace
Em = ½ m (2π r / T)² + U
Em = 2π² m r² / T² + U
Let's calculate
Em = 2π² 425 (2.01 10⁷)² / (4.2120 10⁴)² - 8.43 10⁹
Em = 1,910 10⁹ - 8.43 10⁹
Em = - 6.52 10⁹ J
Answer:
a) [tex]-8.44 * 10^9[/tex] J
b) [tex]2.24 * 10^{10}[/tex] J
Explanation:
Given and known parameters are:
Gravitational constant, [tex]G = 6.674*10^{-11}[/tex] [tex]N.(m/kg)^2[/tex]
Mass of Earth, [tex]M = 5.98 * 10^{24}[/tex] kg
Mass of satellite, m = 425 kg
Radius if Earth, R = 6370 km
Radius of satellite, r = 20100 km
Period, T = 702 min
a) Potential energy in the satellite's orbit is:
[tex]PE = -\frac{GMm}{r} = -\frac{6.674*10^{-11}*5.98 * 10^{24}*425}{20100*10^3} = -8.44 * 10^9[/tex] J
b) To find the energy required to place the satellite in orbit around Earth, we need to determine kinetic energy KE and potential energy PE both initially at rest and finally in the orbit.
[tex]PE_i = -\frac{GMm}{R}\\KE_i = 0\\\\PE_f = -\frac{GMm}{r}\\KE_f = \frac{GMm}{2r}[/tex]
Hence,
[tex]\Delta E = (KE + PE)_{final} - (KE+PE)_{initial} = (-\frac{GMm}{r}+-\frac{GMm}{2r}) - (\frac{GMm}{R})[/tex]
[tex]= GMm(\frac{1}{R}-\frac{1}{2r}) = 6.674*10^{-11}*5.98 * 10^{24}*425*(\frac{1}{6.37*10^6}-\frac{1}{2*20.1*10^6})[/tex]
[tex]= 2.24 * 10^{10}[/tex] J
