Calculate the gravitational force between Earth (M_E = 5.97 ร 10ยฒโด kg) and the Moon (M_M = 7.35 ร 10ยฒยฒ kg), separated by r = 3.84 ร 10โธ m. Use G = 6.67 ร 10โปยนยน Nยทmยฒ/kgยฒ.
Newton's law of gravitation, orbital mechanics, and Kepler's laws
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1
โ
m2โ
โ
where G=6.67ร10โ11 Nยทmยฒ/kgยฒ is gravitational constant.
Vector form:F12โ=โGr2m1โm2โโr^12โ
Gravitational Potential Energy
Choosing U=0 at r=โ:
U(r)=โGrm1โm2โโ
Work to bring masses from infinity to separation r:
W=โซโrโFdrโฒ=โGrm1โm2โโ
Gravitational Field
gโ=โr2GMโr^
Gravitational potential:ฮฆ=โrGMโ
gโ=โโฮฆ
Escape Velocity
Minimum velocity to escape gravitational field:
Set total energy = 0:
21โmvesc2โโGRMmโ=0
vescโ=R2GMโโ
For Earth: vescโโ11.2 km/s
Circular Orbits
For circular orbit of radius r:
Centripetal force = Gravitational force:
rmv2โ=Gr2Mmโ
Orbital velocity:v=rGMโโ
Orbital period:T=v2ฯrโ=2ฯGMr3โโ
Orbital energy:E=KE+PE=21โmv2โGrMmโ
E=โ2rGMmโ
(Negative: bound orbit)
Note: โฃEโฃ=KE (virial theorem)
Kepler's Laws
First Law (Law of Ellipses):
Planets move in elliptical orbits with the Sun at one focus.
Second Law (Law of Equal Areas):
Line from Sun to planet sweeps equal areas in equal times.
This follows from angular momentum conservation:
dtdAโ=2mLโ=constant
A satellite orbits Earth at altitude h = 400 km above the surface. Given R_E = 6.37 ร 10โถ m and g = 9.8 m/sยฒ at surface, find: (a) the orbital speed, (b) the orbital period, and (c) the satellite's acceleration.
๐ก Show Solution
Given:
h = 400 km = 4.0 ร 10โต m
R_E = 6.37 ร 10โถ m
g = 9.8 m/sยฒ
Orbital radius: r=REโ+h=6.77ร106 m
(a) Orbital speed:
For circular orbit:
rmv2โ=
v=rGMEโโ
Using g=RE2โGM โ :
v=rgRE
v=9.8โโ
v=3.13ร2.60ร1036.37ร10
v=7670ย m/s=7.67ย km/sโ
(b) Orbital period:
T=v2ฯrโ=
T=76704.25ร107โ=5546ย s
T=92.4ย min=1.54ย hoursโ
(c) Acceleration:
Centripetal acceleration:
a=rv2โ=
a=6.77ร1065.88ร107โ
a=8.69ย m/s2โ
This is 89% of g at surface (slightly weaker due to altitude).
3Problem 3hard
โ Question:
Derive the escape velocity from Earth's surface using energy conservation. Then calculate the numerical value for Earth (M_E = 5.97 ร 10ยฒโด kg, R_E = 6.37 ร 10โถ m). Also find the escape velocity from the Moon (M_M = 7.35 ร 10ยฒยฒ kg, R_M = 1.74 ร 10โถ m).
๐ก Show Solution
Derivation:
At surface: Eiโ=KE+PE=21โmv2โREโGMEโmโ
At infinity (just escaping): Efโ=0+0=0
Energy conservation: Eiโ=Efโ
21โmvesc2โโ
vescโ=RE
Using g=GMEโ/RE2โ:
vescโ=2gRE
From Earth:
vescโ=2(9.8)(6.37ร106)
vescโ=1.25ร108
vesc,Earthโ=11,200ย m/s=
From Moon:
Surface gravity on Moon:
gMโ=R
gMโ=1.62ย m/s2
vesc,Moonโ=2(1.62)(1.74ร10
vesc,Moonโ=5.64ร106
vesc,Moonโ=2370ย m/s=2.37ย km/s
Moon's lower mass and smaller radius make escape much easier!
Fun fact: This is why Moon has no atmosphere - gas molecules can reach v_esc and escape.
โพ
Yes, this page includes 3 practice problems with detailed solutions. Each problem includes a step-by-step explanation to help you understand the approach.