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
Rom4ik [11]
3 years ago
9

Calculate how much work is required to launch a spacecraft of mass m from the surface of the earth (mass mE, radius RE) and plac

e it in a circular low earth orbit--that is, an orbit whose altitude above the earth's surface is much less than RE. (As an example, the International Space Station is in low earth orbit at an altitude of about 400 km, much less than RE = 6370 km.) Ignore the kinetic energy that the spacecraft has on the ground due to the earth's rotation.
Physics
1 answer:
mylen [45]3 years ago
3 0

To solve this problem it is necessary to apply the concepts related to the conservation of energy, through the balance between the work done and its respective transformation from the gravitational potential energy.

Mathematically the conservation of these two energies can be given through

W = U_f - U_i

Where,

W = Work

U_f = Final gravitational Potential energy

U_i = Initial gravitational Potential energy

When the spacecraft of mass m is on the surface of the earth then the energy possessed by it

U_i = \frac{-GMm}{R}

Where

M = mass of earth

m = Mass of spacecraft

R = Radius of earth

Let the spacecraft is now in an orbit whose attitude is R_{orbit} \approx R then the energy possessed by the spacecraft is

U_f = \frac{-GMm}{2R}

Work needed to put it in orbit is the difference between the above two

W = U_f - U_i

W = -GMm (\frac{1}{2R}-\frac{1}{R})

Therefore the work required to launch a spacecraft from the surface of the Eart andplace it ina circularlow earth orbit is

W = \frac{GMm}{2R}

You might be interested in
On Mars, the acceleration due to gravity is 3.77 m / s 2 . 3.77 m/s2. How far would a 15 g 15 g rock fall from rest in 2.5 s 2.5
Zinaida [17]

Answer:

s = 11.78 m

Explanation:

given,

acceleration due to gravity, g = 3.77 m/s²

mass of the rock = 15 g

time = 2.5 s

distance traveled = ?

using equation of motion

s = ut +\dfrac{1}{2}at^2

initial speed = 0 m/s

s = \dfrac{1}{2}at^2

s = \dfrac{1}{2}\times 3.77 \times 2.5^2

s = 11.78 m

distance traveled by the rock is equal to 11.78 m.

7 0
3 years ago
Read 2 more answers
Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer radius R2. The electric field at a po
gregori [183]

Answer:

E = 0    r <R₁

Explanation:

If we use Gauss's law

      Ф = ∫ E. dA = q_{int} / ε₀

in this case the charge is distributed throughout the spherical shell and as we are asked for the field for a radius smaller than the radius of the spherical shell, therefore, THERE ARE NO CHARGES INSIDE this surface.

Consequently by Gauss's law the electric field is ZERO

           E = 0    r <R₁

6 0
3 years ago
How much force is needed to stop a 4000 kg truck moving at 8 m/s in 0 2 seconds?​
maksim [4K]

Answer:

32000 N

Explanation:

Force Calculater

8 0
3 years ago
10.   Cumulus and stratus clouds belong to which cloud group? <br><br>     
EastWind [94]
Cumulus belongs to vertical clouds and status to low
5 0
4 years ago
A small branch is wedged under a 200 kg rock and rests on a smaller object. The smaller object is 2.0 m from the large rock and
Alexxandr [17]

Answer:

a

  F  =326.7 \ N

b

  M  = 6

Explanation:

From the question we are told that

          The mass of the rock is  m_r  =  200 \ kg

          The  length of the small object from the rock is  d  =  2 \ m

          The  length of the small object from the branch l  =  12 \ m

An image representing this lever set-up is shown on the first uploaded image

Here the small object acts as a fulcrum

The  force exerted by the weight of the rock is mathematically evaluated as

      W =  m_r *  g

substituting values

     W =   200 *  9.8

     W =   1960 \ N

 So  at  equilibrium the sum  of the moment about the fulcrum is mathematically represented as

         \sum  M_f  =  F * cos \theta *  l  -  W cos\theta  *  d =  0

Here  \theta is very small so  cos\theta  *  l  =  l

                               and  cos\theta  *  d  =  d

Hence

       F *   l  -  W  * d =  0

=>    F  = \frac{W * d}{l}

substituting values

        F  = \frac{1960 *  2}{12}

       F  =326.7 \ N

The  mechanical advantage is mathematically evaluated as

          M  = \frac{W}{F}

substituting values

        M  = \frac{1960}{326.7}

       M  = 6

6 0
3 years ago
Other questions:
  • Two thermally insulated cylinders, A and B, of equal volume, both equipped with pistons, are connected by a valve. Initially A h
    15·1 answer
  • Consider a keen little boy who is having a wagon race with a friend. He starts from rest and
    11·1 answer
  • Which of the following structure's primary role is to filter lymph?
    7·1 answer
  • To practice Problem-Solving Strategy 12.1 for rotational dynamics problems. Suppose that you are holding a pencil balanced on it
    7·1 answer
  • Most stars are<br> a.brown fs<br> b.main sequence<br> c.white dwarfs<br> d.pulsars
    15·1 answer
  • If position vector r = bt^2i + ct^3j, where b and c are positive constants, when does the velocity vector make an angle of 450 w
    10·1 answer
  • A flower is red because..?<br><br> Please help! Thank you! I need to get this done ASAP!?!!?
    5·1 answer
  • What are the two sources of earth's internal heat energy
    7·1 answer
  • A elephant kicks a 5.0 kg stone with 150 J of kinetic energy.
    11·1 answer
  • How much work does it take to accelerate from 5 m/s to 10 m/s?
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!