1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Slav-nsk [51]
3 years ago
9

The acceleration due to gravity, g , is constant at sea level on the Earth's surface. However, the acceleration decreases as an

object moves away from the Earth's surface due to the increase in distance from the center of the Earth. Derive an expression for the acceleration due to gravity at a distance h above the surface of the Earth, g h . Express the equation in terms of the radius R of the Earth, g , and h .
Physics
1 answer:
blsea [12.9K]3 years ago
3 0

Answer:

  g    = g₀   [1- 2 h / Re + 3 (h / Re)²]

Explanation:

The law of universal gravitation is

        F = G m Me / Re²

Where g is the universal gravitational constant, m and Me are the mass of the body and the Earth, respectively and R is the distance between them

      F = G Me /Re²  m

We call gravity acceleration a

       g₀ = G Me / Re².

When the body is at a height h above the surface the distance is

            R = Re + h

Therefore  the attractive force is

      F = G Me m / (Re + h)²

Let's take Re's common factor

      F = G Me / Re²  m / (1+ h / Re)²

As Re has a value of 6.37 10⁶ m and the height of the body in general is less than 10⁴ m, the h / Re term is very small, so we can perform a series expansion

         (1+ h / Re)⁻² = 1 -2 h / Re + 6/2 (h / Re) 2 + ...

Let's replace

       F = G Me /Re²   m [1- 2 h / Re + 3 (h / Re)²]

       F = g₀   m  [1- 2 h / Re + 3 (h / Re)²]

If we call the force of attraction at height

     m g =g₀ m  [1- 2 h / Re + 3 (h / Re)²]

       g    = g₀   [1- 2 h / Re + 3 (h / Re)²]

You might be interested in
Name each type of symbiosis and explain how the two species are affected
disa [49]
Frogs
snakes if there food chain is mesesed up it dont work no more

7 0
3 years ago
Suppose you observe that at night, the air just above the ground feels cooler than the air above it. then in the middle of a sun
yanalaym [24]
The answer is convection
6 0
3 years ago
Compare the time period of two simple pendulums of length 4m and 16m at a place.
Vlad1618 [11]

Answer:

the period of the 16 m pendulum is twice the period of the 4 m pendulum

Explanation:

Recall that the period (T) of a pendulum of length (L)  is defined as:

T=2\,\pi\,\sqrt{ \frac{L}{g} }

where "g" is the local acceleration of gravity.

SInce both pendulums are at the same place, "g" is the same for both, and when we compare the two periods, we get:

T_1=2\,\pi\,\sqrt{\frac{4}{g} } \\T_2=2\,\pi\,\sqrt{\frac{16}{g} } \\ \\\frac{T_2}{T_1} =\sqrt{\frac{16}{4} } =2

therefore the period of the 16 m pendulum is twice the period of the 4 m pendulum.

5 0
2 years ago
The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit
vovikov84 [41]

Gravitational force is given by, F= G\frac{mM}{R^{2}}

Where, m and M are the masses of the objects, R is the distance between them and G gravitational constant.

Gravitational force of the star on planet 1, F_{1}= G\frac{m_{1}M}{R^{2}}

Gravitational force of the star on planet 2, F_{2}= G\frac{3m_{1}M}{(3R)^{2}}

Ratio, \frac{F_{1}}{F_{2}}= \frac{\frac{Gm_{1}M}{R^{2}}}{\frac{G3m_{1}M}{(3R)^{2}}}

\frac{F_{1}}{F_{2}}=  \frac{3}{1}

Therefore, the gravitational force of the star on the planet 1 is three times that on planet 2.

6 0
3 years ago
Read 2 more answers
An 8.5 kg crate is pulled 5.1 m up a 30 degree incline by a rope angled 17 degrees above the incline. The tension in the rope is
Ulleksa [173]

Answer:

1. a W_t=746.63 J

  b E_p=212.415 J

  c W_n=183.96J

2. T_e=99.71J

Explanation:

a). The work done by the tension is:

W_t=T*dt

dt=\frac{5.1}{cos(17)}

W_t=140N*\frac{5.1}{cos(17)}

W_t=746.63 J

b). The work done potential of gravity

E_p=m*g*h

h=5.1*sin(17)

E_p=8.5kh*9.8*5.1*sin(30)

E_p=212.415 J

c). The work done by the normal force

W_n=N*d_n

d_n=5.1*sin(30)=2.55

W_n=8.5kg*9.8*cos(30)*2.55

W_n=183.96J

2. The increase in thermal energy is:

T_e=F*d

F_k=u_k*m*g=0.271*8.5kg*9.8*cos(30)

F_k=19.5N

T_e=19.55*5.1m

T_e=99.71J

4 0
3 years ago
Other questions:
  • List some things in your house that has the same density....
    13·1 answer
  • Now assume that the water is contained in a 0.1-kg aluminum pot (cAl = 900 J/kg-K) that is initially at 293 K just like the wate
    11·1 answer
  • A 3.1 kg dog stands on an 18 kg flatboat and is 6.1 m from the shore. He walks 2.5 m on the boat toward shore and then stops. As
    5·1 answer
  • You are looking at your textbook. you can see the textbook because of light doing what process when it hits the textbook's surfa
    6·1 answer
  • A hole is to be drilled in the plate at A. The diameters of the bits available to drill the hole range from 12 to 24 mm in 3-mm.
    12·1 answer
  • The coefficient of static friction between hard rubber and normal street pavement is about 0.90. On how steep a hill (maximum an
    14·2 answers
  • Un avión tarda en llegar a su destino 12 horas. Si recorrió una distancia de 10.700 kilómetros. Calcular su velocidad y expesarl
    10·1 answer
  • Cylindrical rod has equal and opposite forces applied perpendicular to its circular ends. The forces are directed away from the
    10·1 answer
  • Which is the distance of the satellite from Earth?
    10·2 answers
  • hai điện tích điểm bằng nhau có độ lớn 6.10^-8C đặt cách nhau một khoảng là r trong dầu hỏa có hằng số điện môi E=2,1. Lực tương
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!