9.3. Circular Orbits http://www.ck12.org
Rs=r−Re= 42 , 227 km− 6 , 378 km= 35 , 849 km→ 3. 58 × 104 kmIllustrative Example 3
a. Exoplanet Gliese 1214b has a mass 6.35 times greater than Earth’s and a radius 2.78 greater than Earth’s. What is
the gravitational acceleration at the surface of Gliese 1214b?
Answer:We use Newton’s Universal Law of Gravity.
We know that the acceleration at the surface of the Earth isge=GmRe 2 e, and therefore, in general, the acceleration of
gravity on the surface of any planet isg=GmR 2. In this case, the massmand the radiusRis of Gliese 1214b. We are
given thatmgliese= 6. 35 meandRgliese= 2. 78 Re.
Substituting, we haveggliese=G( 2 (.^678.^35 Rme)e 2 )= 0. 8216 GmR (^2) ee→ 0. 8216 ( 9. 81 ) = 8. 059 → 8. (^06) sm 2.
b. What would a person who weighs 700 N on Earth weigh on Gliese 1214b?
Answer:Let’s solve this two ways.
i. Use the person’s mass and multiply it by the acceleration of gravity on Gliese 1214b.
The mass of the person ism=Wge= 9700. 81 Nm
s^2
= 71. 356 kg. Therefore, the weight of the person on Gliese 1214b
isW=mg= ( 71. 36 )( 8. 06 ) = 575. 1 → 575 N
ii. Find the ratio of the gravitational acceleration on Gliese compared to the Earth, and multiply the weight of the
person on Earth by that ratio.
8. 06
9. 81 (^700 ) =^575.^1 →^575 N
- Kepler’s Three Laws of Planetary Motion are:
i. The orbital paths of the planets about the sun are ellipses with the sun at one focus.
ii. If an imaginary line is drawn from the sun to a planet as the planet orbits the sun, this line will sweep
out equal areas in equal times. (A planet moves faster when it is closer to the sun and slower when it is
farther away from the sun.)
iii. The square of the timeT^2 for the orbital period of a planet about the sun is proportional to the cube of
the average distancer^3 between the sun and the planet. That is,T^2 ∝r^3 orT^2 =kr^3 wherekequals^4 π
2
GM
andMis the mass of the central body
(
T^2 =^4 π
2
Gmr
3)
. IfTis expressed in years andrin astronomical
units thank=1 andT^2 =r^3
2.The Universal Law of GravityThe forceFbetween two objects is directly proportional to the product of
their masses,m 1 m 2 , and inversely proportional to the square of the distance,r^2 between their centers:
F=
Gm 1 m 2
r^2whereGisthe universal gravitational constantequal toG= 6. 67 × 10 −^11 N·m
2
kg^2.- The gravitational acceleration near a massive body of massmisg=Gmr 2 wheremis the mass that creates the
gravitational acceleration andris the distance from the center of the planet to a point outside the planet. - The electrostatic force between two charged bodies isF=kqr^12 q^2.