SME
Classical Mechanics · Problem type
Circular and escape speed
Use this for “How fast must a probe move to circle Mars (GM = 4.28 × 10¹³ m³/s²) 3.6 × 10⁶ m from its center, and how fast to escape from there?”
Diagram
Orbital
speed:
v
=
√(
μ
r
)
=
√(
3.986 ×
10
14
6.371
×
10
6
)
=
7,910
m/s
Period:
T
=
2π√(
r
3
μ
)
=
2π
×
√(
6.371 ×
10
63
3.986
×
10
14
)
=
5,060
s
(84.3
min)
Gravity
is
the
centripetal
force:
μ
r
2
=
v
2
r
,
so
the
satellite’s
own
mass
cancels.
This is an example. Type your own values to start a new problem.
v
c
=
GM (m³/s²)
r (m)
=
vc (m/s)
v
esc
=
2
×
vc
=
vesc (m/s)
M
Central mass
worked out (not typed)
Formulas
vc
=
√(
GM
r
)
7,910
=
√(
3.986 ×
10
14
6.371 ×
10
6
)
vesc
=
√
2
×
vc
11,190
=
√
2
×
7,910
M
=
GM
6.674 ×
10
-11
5.972 ×
10
24
=
3.986 ×
10
14
6.674 ×
10
-11
Assumptions
•
One body far heavier than the other; r is measured from its center. Earth: GM = 3.986 × 10¹⁴ m³/s², R = 6.371 × 10⁶ m.
•
On a circle gravity supplies the centripetal force, GM ÷ r² = vc² ÷ r.
•
Escape: ½v² − GM ÷ r = 0, the energy of a parabola, so vesc² = 2GM ÷ r; no air drag. G = 6.674 × 10⁻¹¹ N·m²/kg².
•
Speeds far below light: at 0.1c the Newtonian answer is still within 1%.
Step-by-step
WE KNOW
GM = 3.986 × 10¹⁴ m³/s², r = 6.371 × 10⁶ m
FIND
circular speed (vc), escape speed (vesc), central mass (M)
Step 1 · Find circular speed (vc)
Use vc = √(GM ÷ r)
Gravity
is
the
centripetal
force:
GM
r
2
=
vc
2
r
,
so
vc
2
=
GM
r
.
vc
=
√(
3.986 ×
10
14
6.371 ×
10
6
)
vc
=
√
6.256 ×
10
7
vc = 7,910 m/s
Step 2 · Find escape speed (vesc)
Use vesc = √2 × vc
Set
1
2
v
2
=
GM
r
:
vesc
2
=
2G
M
r
,
twice
vc
2
,
so
vesc
=
√
2
×
vc.
vesc
=
√
2
×
7,909.788
vesc = 1.4142 × 7,909.788
vesc = 11,190 m/s
Step 3 · Find central mass (M)
Use M = GM ÷ (6.674 × 10⁻¹¹)
GM
is
known
far
better
than
G
or
M
alone;
dividing
by
G
gives
the
mass.
M
=
3.986 ×
10
14
6.674 ×
10
-11
M
=
5.972 ×
10
24
kg
Check
Put the numbers back in. Both sides should match.
7,910
=
√(
3.986 ×
10
14
6.371 ×
10
6
)
✓
11,190
=
√
2
×
7,910
✓
5.972 ×
10
24
=
3.986 ×
10
14
6.674 ×
10
-11
✓
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An orbit’s turning points from its energy
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A Hohmann transfer between circular orbits
Topic check
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Related lessons
Central forces
Main lesson
An orbit’s turning points from its energy
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Use this for “A satellite in a circular orbit 7000 km from Earth’s center moves to one at 26,600 km on a Hohmann transfer. Find both burns, the total Δv and the coast time.”