Gravitational Acceleration and Density Calculations for Various Planets and Objects, Assignments of Astronomy

Solutions to problem set 1 of the astro/eps c12 (2008) course, focusing on calculating surface gravity and gravitational acceleration for various planets, objects, and the international space station. It also includes calculations for the density of enchilladus and an analysis of its composition.

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Pre 2010

Uploaded on 10/01/2009

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Astro/EPS C12 (2008) Mike Wong
Problem Set 1 Solution Guide
1 Surface gravity
Problem 1A
First find a formula for g:
F=mobject g
F=G mplanet mobject
r2
mobject g=G mplanet mobject
r2
g=G mplanet
r2
For the last column of the table, just divide by the Earth’s surface gravitational
acceleration g= 9.8 m sec2.
As an example, here is the calculation for the International Space Station. The
other parts of the problem are done the same way. Use the distance to the
Earth’s center (350 km + the Earth’s radius) and the mass of the Earth, be-
cause the space station itself is too tiny to exert much of a gravitational pull on
anything.
r=r+ 350 km = 6370 km + 350 km = 6720 km ×1000 m
1 km = 6.72 ×106m
mplanet = 5.97 ×1024 kg
g=G mplanet
r2=6.67 ×1011 m3kg1sec25.97 ×1024 kg
(6.72 ×106)2m2= 8.82 m sec2
% of Earth surface gravity = g/g= 8.82 m sec2/9.8 m sec2= 90%
1
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Astro/EPS C12 (2008) – Mike Wong Problem Set 1 Solution Guide

1 Surface gravity

Problem 1A

First find a formula for g:

F = mobject g

F =

G mplanet mobject r^2

mobject g =

G mplanet mobject r^2

g = G mplanet r^2

For the last column of the table, just divide by the Earth’s surface gravitational acceleration g⊕ = 9.8 m sec−^2.

As an example, here is the calculation for the International Space Station. The other parts of the problem are done the same way. Use the distance to the Earth’s center (350 km + the Earth’s radius) and the mass of the Earth, be- cause the space station itself is too tiny to exert much of a gravitational pull on anything.

r = r⊕ + 350 km = 6370 km + 350 km = 6720 km ×

1000 m 1 km

= 6. 72 × 106 m

mplanet = 5. 97 × 1024 kg

g =

G mplanet r^2

  1. 67 × 10 −^11 m^3 kg−^1 sec−^2 5. 97 × 1024 kg (6. 72 × 106 )^2 m^2

= 8.82 m sec−^2

% of Earth surface gravity = g/g⊕ = 8.82 m sec−^2 / 9 .8 m sec−^2 = 90%

Location

Distance to center of planet/object (m)

Mass of planet/ object (kg)

Gravitational acceleration

(m s–2)

(% of Earth surface gravity) Surface of the Moon 1.74 106 7.35 1022 1.6 17% Surface of Pluto 1.18 106 (p.75) 1.3 1022 0.62 6.4% Surface of Ceres 4.57 105 9.4 1020 0.3 3% Surface of Enchilladus 5.4 105 1.1 1021 0.25 2.6% International Space Station orbiting 350 km above Earth’s surface

6.7 106 5.97 1024 8.8 90%

Problem 1B “Weight is a measure of gravitational force” means we can use the F from part 1A as the weight. Then:

F⊕ = mobject g⊕ FPluto = mobject gPluto

FPluto F⊕

mobject gPluto mobject g⊕

FPluto = F⊕ gPluto g⊕

= 150 lbs × 6 .4% = 9.53 lbs

Problem 1C Solve the equation for g:

g = F/m = 150 lbs/150 lbs = 1

So the answer is just g = 1, which is valid only at the Earth’s surface. So g is actually a convenient unit of acceleration, used for measuring accelerations felt by astronauts during launch, by roller-coaster riders, and by football tacklers.

Problem 1D The astronauts experience a zero-gravity environment because they are in a kind of free-fall. The 0.9 g acceleration at the space station is what keeps it in orbit around the Earth. Astronauts train for zero-g environments by riding in an airplane called the “vomit comet,” which goes up high, then dives down. During the descent, passengers feel weightless because they’re free-falling.