Test #1. Problem 2- Universal gravity (a) E = Mu²- MEMG = (2M-G) 12 Vesc= At lift-off, r=r and E=Mvoc - Mresc Vesc = 1.1 x 10 m/s. + 13 = + ½ 1 Eth Earth radius (c). v = 1 / Vese. Therefore - Mesc E = - 3 x 800 kg x (1.1) 10 m² YE = 43 I h = 1/16 2100 km When the rocket reaches maximum altitude, Tmax = rεth = 1/1 1E and v = 0. In order escape to from planet Earth, the pre- programmed explosion must impart a vebuity, Veap, to the part with mass MA= 1M, such that E = MV- MEMG rmax 0 = (b) 22 3.6 × 100 J At maximum altitude, v=o and rreth. Therefore max -2- 2 Therefore: Ve=Vesc This velocity is directly radially outward from planet Earth. Now, the other half ice,in 19:34 Sun 23 Jan ,in=20°C O toms ← 22/01/2024 Problem 2-Universal gravity and conservation of energy A rocket is launched vertically from the surface of the Earth with an initial velocity Vyo = 0.5 Vesc, where Vesc is the escape velocity from the surface of planet Earth. No further thrust is imparted to the rocket after lift-off. The rocket has a mass M-800 kg. Neglect air drag. (a) Find the value of the constant mechanical energy (Hint: use Vesc = 1.1x104m/s). (b) Find the maximum altitude h (in km) above the Earth surface reached by the rocket. (c) Suppose, now, that when the rocket reaches maximum altitude, a pre-programmed explosion separates the rocket in two parts of equal mass. One part is projected radially outward, away from the Earth, reaching an infinite distance from Earth. Evaluate the minimum energy released by the pre-programmed explosion, assuming that this energy is entirely transformed into the total kinetic energy of the two parts. (a) 1) Vyo = 1 0.5 Vesc M-800kg E=Mv²- GMEM , r VESC = 2MEG Vesc=11x10 E-(u)² - GHEM 1M ves - GHEM Mvc - GM) M (1½ v/ SC - ½ Ve³c) = -3 Musc E = -3x 800 x (1.1 x 10")² = -3.6x100 (b) max = reth E. mv=GMEM Tmax >h= v² GME MYE+h = 2GME >h= 2G-ME 2GME Vese) v² Eth -YE h=4YE-YE MEG YE Paste 2G-ME -YE v²
Test #1. Problem 2- Universal gravity (a) E = Mu²- MEMG = (2M-G) 12 Vesc= At lift-off, r=r and E=Mvoc - Mresc Vesc = 1.1 x 10 m/s. + 13 = + ½ 1 Eth Earth radius (c). v = 1 / Vese. Therefore - Mesc E = - 3 x 800 kg x (1.1) 10 m² YE = 43 I h = 1/16 2100 km When the rocket reaches maximum altitude, Tmax = rεth = 1/1 1E and v = 0. In order escape to from planet Earth, the pre- programmed explosion must impart a vebuity, Veap, to the part with mass MA= 1M, such that E = MV- MEMG rmax 0 = (b) 22 3.6 × 100 J At maximum altitude, v=o and rreth. Therefore max -2- 2 Therefore: Ve=Vesc This velocity is directly radially outward from planet Earth. Now, the other half ice,in 19:34 Sun 23 Jan ,in=20°C O toms ← 22/01/2024 Problem 2-Universal gravity and conservation of energy A rocket is launched vertically from the surface of the Earth with an initial velocity Vyo = 0.5 Vesc, where Vesc is the escape velocity from the surface of planet Earth. No further thrust is imparted to the rocket after lift-off. The rocket has a mass M-800 kg. Neglect air drag. (a) Find the value of the constant mechanical energy (Hint: use Vesc = 1.1x104m/s). (b) Find the maximum altitude h (in km) above the Earth surface reached by the rocket. (c) Suppose, now, that when the rocket reaches maximum altitude, a pre-programmed explosion separates the rocket in two parts of equal mass. One part is projected radially outward, away from the Earth, reaching an infinite distance from Earth. Evaluate the minimum energy released by the pre-programmed explosion, assuming that this energy is entirely transformed into the total kinetic energy of the two parts. (a) 1) Vyo = 1 0.5 Vesc M-800kg E=Mv²- GMEM , r VESC = 2MEG Vesc=11x10 E-(u)² - GHEM 1M ves - GHEM Mvc - GM) M (1½ v/ SC - ½ Ve³c) = -3 Musc E = -3x 800 x (1.1 x 10")² = -3.6x100 (b) max = reth E. mv=GMEM Tmax >h= v² GME MYE+h = 2GME >h= 2G-ME 2GME Vese) v² Eth -YE h=4YE-YE MEG YE Paste 2G-ME -YE v²
College Physics
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Chapter1: Units, Trigonometry. And Vectors
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I keep getting stuck in part b could you show me how to do part b
Transcribed Image Text:Test #1. Problem 2- Universal gravity
(a) E = Mu²- MEMG
= (2M-G) 12
Vesc=
At lift-off, r=r and
E=Mvoc - Mresc
Vesc = 1.1 x 10 m/s.
+
13 = + ½ 1
Eth
Earth radius
(c).
v = 1 / Vese. Therefore
- Mesc
E = - 3 x 800 kg x (1.1) 10 m²
YE
=
43
I
h = 1/16
2100 km
When the rocket reaches maximum altitude,
Tmax = rεth = 1/1 1E and v = 0. In order
escape
to from planet Earth, the pre-
programmed explosion must impart a vebuity,
Veap, to the part with mass MA= 1M, such that
E = MV-
MEMG
rmax
0
=
(b)
22
3.6 × 100 J
At maximum altitude, v=o and
rreth. Therefore
max
-2-
2
Therefore:
Ve=Vesc
This velocity is directly radially outward
from planet Earth. Now, the other half
Transcribed Image Text:ice,in
19:34 Sun 23 Jan
,in=20°C
O
toms
←
22/01/2024
Problem 2-Universal gravity and conservation of energy
A rocket is launched vertically from the surface of the Earth with an initial velocity Vyo = 0.5 Vesc, where Vesc is
the escape velocity from the surface of planet Earth. No further thrust is imparted to the rocket after lift-off. The
rocket has a mass M-800 kg. Neglect air drag.
(a) Find the value of the constant mechanical energy (Hint: use Vesc = 1.1x104m/s).
(b) Find the maximum altitude h (in km) above the Earth surface reached by the rocket.
(c) Suppose, now, that when the rocket reaches maximum altitude, a pre-programmed explosion separates
the rocket in two parts of equal mass. One part is projected radially outward, away from the Earth, reaching
an infinite distance from Earth. Evaluate the minimum energy released by the pre-programmed explosion,
assuming that this energy is entirely transformed into the total kinetic energy of the two parts.
(a)
1) Vyo = 1
0.5 Vesc
M-800kg
E=Mv²- GMEM
,
r
VESC =
2MEG
Vesc=11x10 E-(u)² - GHEM 1M ves - GHEM
Mvc - GM) M (1½ v/ SC - ½ Ve³c) = -3 Musc
E = -3x 800 x (1.1 x 10")² = -3.6x100
(b) max = reth E.
mv=GMEM
Tmax
>h=
v² GME MYE+h = 2GME >h=
2G-ME 2GME
Vese)
v²
Eth
-YE
h=4YE-YE
MEG
YE
Paste
2G-ME
-YE
v²
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