Gravitation Earth’s Surface (Gravitation Field) Problems and Solutions

 Problem#1

(a) What will an object weigh on the Moon’s surface if it weighs 100 N on Earth’s surface? (b) How many Earth radius must this same object be from the center of Earth if it is to weigh the same as it does on the Moon?

Answer:
(a) the gravitational acceleration at the surface of the Moon is

g_Earth/g_moon = W_Earth/W_moon
with, g_Earth = 9.81 m/s² and g_moon = 1.67 m/s²
then,
[9.81 m/s²/1.67 m/s²] = 100 N/W_moon
W_moon = 17.02 N

(b) If the force f of 17.02 N above is the gravitational force between the object and the earth, we can express that using Newton's universal law of gravity.

F = GmM/r²

With, r = distance from the center of the earth to the object, M = mass of the earth
So, the distance r is

17.02 N = (6.67 × 10⁻¹¹ Nm²/kg²)(10.19 kg)( 5.98 × 10²⁴ kg)/r²
r = 1.545 x 10⁷ m

The radius of the earth is 6378 km, so the number of the earth's radius is
N = 1.545 × 10⁷ m/6378000 m = 2.42

Problem#2
A large mountain can slightly affect the direction of “down” as determined by a plumb line. Assume that we can model a mountain as a sphere of radius R = 2.00 km and density (mass per unit volume) 2.6 x 10³ kg/m³. Assume also that we hang a 0.50 m plumb line at a distance of 3R from the sphere’s center and such that the sphere pulls horizontally on the lower end. How far would the lower end move toward the sphere?

Answer:
Known:
sphere of radius R = 2.00 km = 2.00 x 10³ m
density (mass per unit volume) ρ = 2.6 x 10³ kg/m³

Fig.1

The free body diagram of plumb line will be here, T is the tension on the string and F_g is gravitational force on plumb by the mountain.

Fig.2

when the system is in equilibrium

F_g = T sin θ and T cos θ = mg

Then,
F_g/mg = tan θ

From the Figure 2, tan θ = d/L, because sin θ <<, so that

d = F_gL/mg = GML/gr²
where d is distance moved from lower end.

M = ρV = ρ(4πR³/3)
M = (2.6 x 10³ kg/m³)[4π x (2.00 x 10³ m)³/3] = 8.71 x 10¹³ kg

So,
d = (6.67 × 10⁻¹¹ Nm²/kg²)(8.71 x 10¹³ kg)(0.50 m)/[9.81 m/s² x (3 x 2.00 x 10³ m)²]
d = 7.4 x 10^-6 m = 7.4 μm

Problem#3
At what altitude above Earth’s surface would the gravitational acceleration be 4.9 m/s²?

Answer:
Let a body of mass m be placed on the surface of the Earth:

g = GM/R²

Fig.3

where M is the mass of the Earth, R is the radius of the Earth and G is the gravitational constant. et the body be now placed at a height h above the Earth's surface.

Let the acceleration due to gravity at that position be g’. Then,

g' = GM/(R + h)²

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

g'/g = [R/(R + h)]²

4.9 m/s²/9.81 m/s² = [6.4×10⁶ m/(6.4 × 10⁶ m + h)]²

0.71 (6.4 × 10⁶ m + h) = 6.4 × 10⁶ m

h = 2.655 x 10⁶ m = 2655 km

Problem#4
In 1956, Frank Lloyd Wright proposed the construction of a mile-high building in Chicago. Suppose the building had been constructed. Ignoring Earth’s rotation, find the change in your weight if you were to ride an elevator from the street level, where you weigh 600 N, to the top of the building.

Answer:
Let a body of mass m be placed on the surface of the Earth:

g = GM_E/R²

Difference in acceleration at head and feet is

da_g/dr = ─2GM_E/r³

which implies that the change in weight is

W_to ─ W_bottom ≈ m(da_g)

However. Since W_bottom = GmM_E/R², we have

m(da_g) = ─2GmM_E (dr/R³)
m(da_g) = ─2W_bottom(dr/R) = ─2 x 600 N x 1.61 x 10³ m/6.37 x 10⁶ m
m(da_g) = ─0.303 N

the minus sign indicating that it is a decrease in W

Problem#5
Certain neutron stars (extremely dense stars) are believed to be rotating at about 1 rev/s. If such a star has a radius of 20 km, what must be its minimum mass so that material on its surface remains in place during the rapid rotation?

Answer:
Known:
the angular velocity, ω = 1 rev/s = 2π rar/s

radius, r = 20 km

The gravitational pull is the centripetal force. If gravity is not strong enough the objects will get thrown off the surface, so they must be at least equal to each other.

GmM/r² = mv²/r or

GM/r = v² Where v = ωr
We want to solve for M:

M = ω²r³/G
M = (2π rad/s)²(20 x 10³ m)³/(6.67 × 10⁻¹¹ Nm²/kg²) = 4.74 x 10²⁴ kg