ASTR 1010 – Spring 2014 – Prof. Magnani
Answer Key – Homework 4
Question 1 – Ch. 4; #28
Mass is the quantity of matter in an object and is measured in kilograms. Weight is a
force and so there must be an acceleration on the object for the weight to be defined.
As a force, the units of weight are Newtons (or kg m/s2).
Question 2 – Ch. 4; #30
The comet in the elliptical orbit is in a BOUND orbit and so it goes around the Sun
with a given period (dictated by Kepler’s Third Law) and “returns” continually. The
comet with the hyperbolic orbit is on an UNBOUND trajectory and so it will pass by
the Sun and then continue outwards until it leaves the Solar System. It will never
come back (like your first love).
Question 3 – Ch. 4; #42
Remember that the acceleration at the surface of a planet is sometimes called “g”
instead of “a”.
gearth = GMearth/(Rearth)2 = 9.8 m/s2
gmars = GMmars/(Rmars)2 = G(0.1*Mearth)/[0.5*(Rearth)]2 = [0.1/0.25][GMearth/(Rearth)2]
gmars = 0.4 gearth
When you use the actual numbers instead of our 0.1, 0.5 approximations, you get
0.39 instead of 0.4.
Let’s say your mass is m = 100 kg
Weight = mg
gMars = 0.39(gearth) = 0.39 * 9.8 = 3.8 m/s2
Weight on Mars = 100 * 3.8 = 380 Nt, instead of the 980 Nt you would weigh on
Earth.
Most Hollywood movies ignore the fact that on Mars you would weigh only 40% of
your current weight on Earth.
Question 4 – Ch. 4; #43
v = 2R/period = 2(4.5 x 109 km) / (164.8 years * 3.16 x 107 s/year) = 5.4 km/s
You can also use the formula for circular velocity where R is now the radius of the
orbit (you will need to know the mass of the Sun which is 2 x 1030 kg):
vcirc = [G Msun/R]0.5
vcirc = [(6.67 x 10-11 Nt m2/kg2) *(2 x 1030 kg)/4.5 x 1012 m)]0.5 = 5.4 x 103 m/s = 5.4
km/s
Note that we had to convert the radius of the orbit into meters before
calculating the value of the velocity. Why? Because G = 6.67 x 10-11 Nt
m2/kg2demands that the lengths be entered as meters.
Question 5 – Ch. 4; #44
For part a) you cannot use the same formula as the problem above because a planet
orbiting a more massive star than the Sun at a distance of 1 AU will have a shorter
period than 1 year. Since we don’t know the period (yet), we will have to use the
circular velocity formula which doesn’t have period in it.
vcirc = [G Mstar/R]0.5 = 4.2 x 104 m/s = 42 km/s
For part b) use Newton’s version of Kepler’s Third Law and solve for p:
p2 = 4 pi2 a3 /(G Mstar)
p = 2.2 x 107 s = 0.71 years
Question 6 – Ch. 4; #45
Use vcirc = [G Msun/R]0.5
and solve for Msun:
Msun = (vcirc)2 R/G = (35.03 x 103 m/s)2 (1.082 x 1011 m)/(6.67 x 10-11 m3/(kg s2) =
1.99 x 1030 kg
Question 7 – Ch. 4; #46
Use vesc = [2G Msun/R]0.5
For Earth:
For Asteorid:
vesc(earth)= [2G (Mearth)/( Rearth)]0.5
vesc(asteroid)= [2G (Masteroid)/( Rasteorid)]0.5
Express the Mass and radius of the asteroid in terms of those of the Earth:
vesc(asteroid) = [2G (10-12 Mearth)/(10-4 Rearth)]0.5 =
[10-8 *2G (Mearth)/( Rearth)]0.5 = 10-4 *[2G (Mearth)/( Rearth)]0.5
vesc(asteroid)= 10-4 vesc(earth)
vesc(asteroid)= 10-4 *(11.2 km/s) = 11.2 x 10-4 km/s = 1.12 x 10-3 km/s = 1.12 m/s
You could throw a baseball off the surface of this asteroid and it would leave on a
hyperbolic orbit – however, it would then be captured by the gravitational pull of
the Sun and orbit the Sun (following Kepler’s Third Law).
Question 8 – Ch.4; #48
a) circular
b) It must escape from the Earth’s gravity. At the surface of the Earth (R=6370
km) that would be 11.2 km/s (use the escape velocity formula; the mass of
the Earth is 6.0 x 1024 kg). If you are orbiting the Earth, you are farther from
the center than 6370 km. Let’s say you are orbiting 300 km above the
surface, then your R in the escape velocity formula is 6370 + 300 = 6670 km.
Plug that into the escape velocity and you get 10.9 km/s as the escape
velocity from a point 300 km above the Earth’s surface.
Question 9 – Ch. 4; #49
Use
a = GM/R2
M = 5.97 x 1024 kg
R = 6370 km =6.37 x 106 m
Plug those values in (along with the value for G = 6.67 x 10-11 Nt m2/kg2) and you
will get 9.81 m/s2