Estimate the mass of Earth  Part 2) The force of Earth's gravitational pull on the moon is about 2.0 x 10^20 N. Estimate the mass of the moon.  Part 3) The moon has a radius of 1.7 x 10^6 m. On the surface of the moon, what is the acceleration due to the moon's gravity?  Part 4) How many g's is your answer to part c? Or what percentage of Earth's gravity would you experience on the moon?  Part 5) If you stood on the moon and jumped straight up with a velocity of 1.0 m/s, how long would it be before

Modern Physics
3rd Edition
ISBN:9781111794378
Author:Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher:Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Chapter4: The Particle Nature Of Matter
Section: Chapter Questions
Problem 6P
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Please refer to the picture for the formulas and constants that have to be used and shown in this problem. 

It takes light about 1.25 seconds to reach the moon from Earth. The moon orbits Earth about once every 27.3 days.

Part 1) Estimate the mass of Earth 

Part 2) The force of Earth's gravitational pull on the moon is about 2.0 x 10^20 N. Estimate the mass of the moon. 

Part 3) The moon has a radius of 1.7 x 10^6 m. On the surface of the moon, what is the acceleration due to the moon's gravity? 

Part 4) How many g's is your answer to part c? Or what percentage of Earth's gravity would you experience on the moon? 

Part 5) If you stood on the moon and jumped straight up with a velocity of 1.0 m/s, how long would it be before you landed back on the moon's surface? 

Please answer all parts. 

 
Please answer in a simply way
Formulas
daug = AV > Vp = √₂ + aaug At
->
At
Vaug = 1/2 (V₂ + √₂)
Vaug = AX X₁ = X₁ + Vaug At
At
X²₁ = X₁ + √₂ At + ½ åst ²
2
2
V² = V₁² + 2aAx
F = ma
F₁ = M₁₂₁ FN
AR = V²
F₁B = - F₁A
BA
F₁ ≤ M₂ FN
FG = G m₁m₂
r²
f ===
T
O
Constants
C≈ 3 x 10³ m/s
9 = 9.8 m/s²
G = 6.7 x 10" Nm² / Kg ²
Units
t = 5
x = m
✓ = m/s
a = m/s²
m = 19
F² = N
T= 3
f = H₂ = 1/s
Geometry
A = lw
A = 1/2 bh
A = πTr ²
C=2 TTC
Tuncertainty Propagation
C=dA →> 0c = |d0A
C = Aª² >> °C = |d² A
Ad
C = A±B20² = 0 ² ² + 0 ²₁₂
ога
C=AB or C => 04
A
न्ह
B
C²
A²
+
B
B²
Transcribed Image Text:Formulas daug = AV > Vp = √₂ + aaug At -> At Vaug = 1/2 (V₂ + √₂) Vaug = AX X₁ = X₁ + Vaug At At X²₁ = X₁ + √₂ At + ½ åst ² 2 2 V² = V₁² + 2aAx F = ma F₁ = M₁₂₁ FN AR = V² F₁B = - F₁A BA F₁ ≤ M₂ FN FG = G m₁m₂ r² f === T O Constants C≈ 3 x 10³ m/s 9 = 9.8 m/s² G = 6.7 x 10" Nm² / Kg ² Units t = 5 x = m ✓ = m/s a = m/s² m = 19 F² = N T= 3 f = H₂ = 1/s Geometry A = lw A = 1/2 bh A = πTr ² C=2 TTC Tuncertainty Propagation C=dA →> 0c = |d0A C = Aª² >> °C = |d² A Ad C = A±B20² = 0 ² ² + 0 ²₁₂ ога C=AB or C => 04 A न्ह B C² A² + B B²
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