All categories

2 years ago

A 4000 kg satellite is placed 2.60 x 10^6 m above the surface of the Earth.

Physics

1 answer:

mash [69]2 years ago

6 0

a) The acceleration of gravity is $4.96 m/s^2$

b) The critical velocity is 6668 m/s (24,006 km/h)

c) The period of the orbit is 8452 s

d) The satellite completes 10.2 orbits per day

e) The escape velocity of the satellite is 9430 m/s

f) The escape velocity of the rocket is 11,191 m/s

Explanation:

The acceleration of gravity for an object near a planet is given by

$g=\frac{GM}{(R+h)^2}$

where

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

h is the height above the surface

In this problem,

$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

$g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)^2}=4.96 m/s^2$

The critical velocity for a satellite orbiting around a planet is given by

$v=\sqrt{\frac{GM}{R+h}}$

where we have again:

$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

$h=2.60\cdot 10^6 m$ (altitude of the satellite)

Substituting,

$v=\sqrt{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=6668 m/s$

Converting into km/h,

$v=6668 m/s \cdot \frac{3600 s/h}{1000 m/km}=24,006 km/h$

The period of the orbit is given by the circumference of the orbit divided by the velocity:

$T=\frac{2\pi (R+h)}{v}$

where

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

v = 6668 m/s

Substituting,

$T=\frac{2\pi (6.37\cdot 10^6 + 2.60\cdot 10^6)}{6668}=8452 s$

One day consists of:

$t = 24 \frac{hours}{day} \cdot 60 \frac{min}{hours} \cdot 60 \frac{s}{min}=86400 s$

While the period of the orbit is

T = 8452 s

So, the number of orbits completed by the satellite in one day is

$n=\frac{t}{T}=\frac{86400}{8452}=10.2$

The escape velocity for an object in the gravitational field of a planet is given by

$v=\sqrt{\frac{2GM}{R+h}}$

where here we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=2.60\cdot 10^6 m$

Substituting, we find

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6 + 2.60\cdot 10^6)}}=9430 m/s$

We can apply again the formula to find the escape velocity for the rocket:

$v=\sqrt{\frac{2GM}{R+h}}$

Where this time we have:

$M=5.98\cdot 10^{24} kg$

$R=6.37\cdot 10^6 m$

$h=0$ , because the rocket is located at the Earth's surface, so its altitude is zero.

And substituting,

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(5.98\cdot 10^{24}}{(6.37\cdot 10^6)}}=11,191 m/s$

Learn more about gravitational force:

brainly.com/question/1724648

brainly.com/question/12785992

#LearnwithBrainly

You might be interested in

Assume a change at the source of sound reduces the wavelength of a sound wave in air by a factor of 3.

noname [10]

Explanation:

The speed of a wave is given by :

$v=f\lambda$ ......(1)

(i) Here, the wavelength of a sound wave in air reduces by a factor of 3. Equation (1) becomes :

$\lambda=\dfrac{v}{f}$

Wavelength and frequency are inversely proportional to each other. So, if wavelength of a sound wave in air reduces by a factor of 3, then the frequency will increases by a factor of 3.

(ii) It remains the same.

8 0

2 years ago

Answer the answers that you know.

sesenic [268]

1). Sequence from the Sun:

   Inner planets:
   Mercury
   Venus
   Earth
   Mars

   Outer planets:
   Jupiter
   Saturn
   Uranus
   Neptune

2). The farther a planet is from the sun, the longer it takes
to orbit the sun. Mercury ... 88 days. Earth ... 365 days.
   Jupiter ... 12 years.    Neptune ... 165 years.

3). Mercury & Venus ... no moons
Earth - 1
Mars - 2
Jupiter - more than 65

4). Mercury ... cratered, no atmosphere
Venus ... cratered, thick cloudy atmosphere
   Mars ... dry, cratered, slight atmosphere, like 1% or Earth's

   Jupiter, Saturn, Uranus, Neptune
   We can't see any surface. If any of them even
   HAS a surface, it's thousands of miles under a
   thick atmosphere of methane gas.

5). Missing from the list

6). Here's a list from the biggest planet to the smallest one.
The numbers in parentheses are the radius of the planet --
half of the diameter:

Jupiter (69,911 km / 43,441 miles) – 1,120% the size of Earth
Saturn (58,232 km / 36,184 miles) – 945% the size of Earth
Uranus (25,362 km / 15,759 miles) – 400% the size of Earth
Neptune (24,622 km / 15,299 miles) – 388% the size of Earth
Earth (6,371 km / 3,959 miles)
Venus (6,052 km / 3,761 miles) – 95% the size of Earth
Mars (3,390 km / 2,460 miles) – 53% the size of Earth
Mercury (2,440 km / 1,516 miles) – 38% the size of Earth

7). At least seven of the planets rotate in the same direction.
There's something different about one of them ... it may be Uranus
but I'm not sure. You'll have to look this up.

8). Saturn has the famous rings, that you can almost see
with only binoculars.
Spacecraft sent to observe the outer planets have detected
very thin rings around Uranus and Neptune.

9). Included in #6.

10). I don't have complete info. Generally, the closer the planet
is to the sun, the hotter it is. But there are a few exceptions.
I think Venus ... the second one from the sun, is actually hotter
than Mercury.

11). Just about every language has its own name for each planet.

12). "Terrestrial" means "like Earth" ("Terra").
The terrestrial planets are the ones that have solid surfaces
and are made of rock.
Mercury, Venus, Earth, and Mars.

13). "Jovian" means "like Jupiter".
Either no solid surface, or very small, inside a big deep gas ball.
Jupiter, Saturn, Uranus, Neptune.

4 0

2 years ago

What does the length of a vector arrow represent?

nalin [4]

The length of a vector arrow represents an magnitude

6 0

1 year ago

During the first 50 s a truck traveled at constant speed of 25 m/s. Find the distance that it is traveled.

Allushta [10]

Time=50s
speed=25m/s

Distance = speed×time
=25×50
=1250m

DISTANCE TRAVELLED IS =1250m

6 0

2 years ago

A proton is projected toward a fixed nucleus of charge Ze with velocity vo. Initially the two particles are very far apart. When

11111nata11111 [884]

Answer:

The value is $R_f = \frac{4}{5} R$

Explanation:

From the question we are told that

The initial velocity of the proton is $v_o$

At a distance R from the nucleus the velocity is $v_1 = \frac{1}{2} v_o$

The velocity considered is $v_2 = \frac{1}{4} v_o$

Generally considering from initial position to a position of distance R from the nucleus

Generally from the law of energy conservation we have that

$\Delta K = \Delta P$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K = K__{R}} - K_i$

=> $\Delta K = \frac{1}{2} * m * v_1^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * (\frac{1}{2} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K = \frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta P = P_f - P_i$

Here $P_i$ is zero because the electric potential energy at the initial stage is zero so

$\Delta P = k * \frac{q_1 * q_2 }{R} - 0$

$\frac{1}{2} * m * \frac{1}{4} * v_o ^2 - \frac{1}{2} * m * v_o^2 = k * \frac{q_1 * q_2 }{R} - 0$

=> $\frac{1}{2} * m *v_0^2 [ \frac{1}{4} -1 ] = k * \frac{q_1 * q_2 }{R}$

=> $- \frac{3}{8} * m *v_0^2 = k * \frac{q_1 * q_2 }{R} ---(1 )$

Generally considering from initial position to a position of distance $R_f$ from the nucleus

Here $R_f$ represented the distance of the proton from the nucleus where the velocity is $\frac{1}{4} v_o$

Generally from the law of energy conservation we have that

$\Delta K_f = \Delta P_f$

Here $\Delta K$ is the change in kinetic energy from initial position to a position of distance R from the nucleus , this is mathematically represented as

$\Delta K_f = K_f - K_i$

=> $\Delta K_f = \frac{1}{2} * m * v_2^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * (\frac{1}{4} * v_o )^2 - \frac{1}{2} * m * v_o^2$

=> $\Delta K_f = \frac{1}{2} * m * \frac{1}{16} * v_o ^2 - \frac{1}{2} * m * v_o^2$

And $\Delta P$ is the change in electric potential energy from initial position to a position of distance $R_f$ from the nucleus , this is mathematically represented as

$\Delta P_f = P_f - P_i$