The Gravitational Field (Weight) Problems and Solutions

 Problem#1

At what distance above the surface of the earth is the  acceleration due to the earth’s gravity  0.980 m/s² if the acceleration due to gravity at the surface has magnitude 9.80 m/s²?

Answer:

An object with mass m will experience a gravitational force equal to

F_g = GmM_E/r²

With r = distance of objects to the center of the earth

Then the gravitational acceleration of the object is given by

a_g = F_g/m = GM_E/r²

Let a body of mass m be placed on the surface of the Earth (r = R_E):

g = GM_E/R_E²

Let the acceleration due to gravity at that position r = R_E + h be g’ . Then,

g' = GM_E/(R_E + h)²

For comparison, the ratio between g’ and g is taken

g'/g = [R_E/(R + h)]²
0.980 m/s²/9.80 m/s² = [6.4×10⁶ m/(6.4 × 10⁶ m + h)]²
0.316 (6.4 × 10⁶ m + h) = 6.4 × 10⁶ m
h = 1.38 x 10⁷ m

Problem#2
The mass of Venus is 81.5% that of the earth, and its radius is 94.9% that of the earth. (a) Compute the acceleration due to gravity on the surface of Venus from these data. (b) If a rock weighs 75.0 N on earth, what would it weigh at the surface of Venus?

Answer:
Known:
Mass of Venus is 81.5% that of the earth, M_V = 0.815M_E

Radius of Venus is 94.9% that of the earth, R_V = 0.949R_E

(a) Let a body of mass m be placed on the surface of the Earth and Venus is

g_E = GM_E/R_E² and g_V = GM_V/R_V²

For comparison, the ratio between g_V and g_E is taken

g_V/g_E = [M_V/M_E][R_E²/R_V²]

g_V/9.80 m/s² = 0.815 x (1/0.949)²

g_V = 8.87 m/s²

(b) If a rock weighs 75.0 N on earth, the mass of the rock is

 m = W_E/g_E = 75.0 N/9.80 m/s² = 7.65 kg

then, the weight of the rock on Venus is 

W_v = mg_V = 7.65 kg x 8.87 m/s² = 67.88 N

Problem#3
Titania, the largest moon of the planet Uranus, has 1/8 the radius of the earth and 1/1700 the mass of the earth. (a) What is the acceleration due to gravity at the surface of Titania? (b) What is the average density of Titania? (This is less than the density of rock, which is one piece of evidence that Titania is made primarily of ice.)

Answer:
Known:
Mass of Titania is 1/1700 that of the Earth, m_T = M_E/1700

Radius of Titania is 1/8 that of the Earth, R_T = R_E/8

the average density of Earth is ρ_{ave, E} = 5.5 g/cm³

(a) the acceleration due to gravity at the surface of Titania is
g_T = GM_T/R_T² = G(M_E/1700)/(R_E/8)²

g_T = (64/1700)GM_E/R_E²

g_T = 0.369 m/s²

(b) the average density of Titania is
ρ_ave,T = m_T/V_T = m_T/(4πR_T³/3)

ρ_ave,T = (M_E/1700)/[4π(R_E/8)³/3]

ρ_ave,T = 512/1700 x ρ_{ave, E} = 1.66 g/cm³

Problem#4 
Rhea, one of Saturn’s moons, has a radius of 765 km and an acceleration due to gravity of 0.278 m/s² at its surface. Calculate its mass and average density.

Answer:
Known:
radius Rhea, R_R = 765 km
gravity of Rhea, g_R = 0.278 m/s²

the acceleration due to gravity at the surface of Rhea is
g_R = GM_R/R_R²

0.278 m/s² = (6.67 × 10⁻¹¹ Nm²/kg²)m_R/(765 x 10³ m)²

m_R = 2.44 x 10²¹ kg

and average density Rhea is
ρ_ave,R = m_R/V_R = m_R/[4π(R_R)³/3]

ρ_ave,R = 3 x 2.44 x 10²¹ kg/[4π x (765 x 10³ m)³]

ρ_ave,R = 1301 kg/m³

Problem#5
Calculate the earth’s gravity force on a 75-kg astronaut who is repairing the Hubble Space Telescope 600 km above the earth’s surface, and then compare this value with his weight at the earth’s surface. In view of your result, explain why we say astronauts are weightless when they orbit the earth in a satellite such as a space shuttle. Is it because the gravitational pull of the earth is negligibly small?

Answer:
the earth’s gravity force on a 75-kg astronaut who is repairing the Hubble Space Telescope 600 km above the earth’s surface is

F’_g = GmM_E/(h + R_E)²

F’_g = (6.67 × 10⁻¹¹ Nm²/kg²)(75 kg)(5.98 × 10²⁴ kg)/(600 x 10³ m + 6378 x 10³ m)²

F’_g = 614.36 N

compare this value with his weight at the earth’s surface, F_g is

F_g = mg = 735 N  

The gravitational pull on the astronaut is therefore not negligible. However, both the
Hubble Space Telescope and the astronaut feel the same acceleration Gm_E/r² and 
therefore don’t accelerate relative to one another, but they also are also on the same orbit
above the earth and are thus constantly “free falling” and do not feel their weight.