Ask question

Ask question

All categories

3 years ago

12

One of your summer lunar space camp activities is to launch a 1130 kg1130 kg rocket from the surface of the Moon. You are a seri

ous space camper and you launch a serious rocket: it reaches an altitude of 215 km215 km . What gain Δ????ΔU in gravitational potential energy does the launch accomplish? The mass and radius of the Moon are 7.36×1022 kg7.36×1022 kg and 1740 km,1740 km, respectively.

1 answer:

maxonik [38]3 years ago

8 0

Answer:

∆U = 2.296×10^10Joules

Explanation:

Gravitational potential energy is defined as the energy possessed by an object under the influence of gravity due to its virtue of position.

Potential energy U = Fr where;

F is the force of attraction between the masses of the moon and the rocket.

r is the radius or height of the object.

From Newton's law of universal gravitation, F = GMm/r²

Potential energy U = (-GMm/r²)×r

Potential energy U = -GMm/r

The force is negative because the objects act upward.

M is the mass of the rocket

m is the mass of the moon

Gravitational potential energy possessed by the rocket

U1 = -GMm/r1

r1 is the altitude covered by the rocket

Gravitational potential energy possessed by the Moon

U2 = -GMm/(r2+r1)

r2 is the radius of the moon

Change in gravitational potential energy ∆U = U2-U1

∆U = -GMm/(r2+r1)-(-GMm/r1)

∆U = -GMm/(r2+r1) + GMm/r1

∆U = -GMm{1/(r2+r1)-1/r1}

Given

G = 6.67×10^-11m³/kgs²

M = 1130kg

m = 7.36×10²²kg

r1 = 215km = 215,000m

r2 = 1740km = 1,740,000m

∆U = -6.67×10^-11× 7.36×10²² × 1130{1/(215,000+1,740,000)-1/215000}

∆U= -55.47×10¹⁴{1/1955000-1/215000}

∆U = -55.47×10¹⁴{5.12×10^-7 - 4.65×10^-6}

∆U = -284×10^7 + 257.94×10^8

∆U = 22,954,000,000Joules

∆U = 2.296×10^10Joules

You might be interested in

An object carries a +15.5 µC charge. It is 0.525 m from a -7.25 µC charge. What is the magnitude of the electric force on the ob

lorasvet [3.4K]

Answer:

the force of attraction between the two charges is 3.55 N.

Explanation:

Given;

first charge carried by the object, q₁ = 15.5 µC

second charge carried by the q₂ = -7.25 µC

distance between the two charges, r = 0.525 m

The force of attraction between the two charges is calculated as;

$F = \frac{kq_1q_2}{r^2} \\\\F = \frac{(9\times 10^9)(15\times 10^{-6})(7.25\times 10^{-6})}{(0.525)^2} \\\\F = 3.55 \ N$

Therefore, the force of attraction between the two charges is 3.55 N.

3 0

2 years ago

When a physical change in a sample occurs what does not change

Lelu [443]

Answer:

When a physical change in a sample occurs, composition of the sample does not change. It stays the same. It stays the same. Also, the properties of the sample will still be the same

5 0

2 years ago

A ball is kicked from the ground into the air at a velocity of 3 m/s at an angle 35° above the horizontal and hits the ground so

Dennis_Churaev [7]

Answer:2.45 m/s

Explanation:

Given

Launch velocity $(u)=3 m/s$

launch angle $=35^{\circ}$

as the vertical velocity first decreasing to zero and then increases to original value so its avg is zero .

$v_{avg}=\frac{displacement}{time}$

$v_{avg}=\frac{Range}{time}$

$v_{avg}=\frac{u\cos \theta \times t}{t}$

thus $v_{avg}=\frac{3\cos 35\times t}{t}$

$v_{avg}=3\cos 35=2.45 m/s$

8 0

3 years ago

The magnitude of the electric field between two parallel charged plates is 200. An electron moves to the negative plate 5. 0 cm

Mama L [17]

The potential difference between the two ends of the circuit is the electric potential difference. The electric potential difference and the work will be 10V and 1.6 x 10^-18 J respectively.

<h3>What is an electric field?</h3>

An electric field is an electric property that is connected with any location in space where a charge exists in any form. The electric force per unit charge is another term for an electric field.

The given data in the problem is given by;

E is the electric field = (200 N/C)

d is the distance = 5.0 cm.=0.05 m

Q is the charge of electrons= 1.602 x 10^-19 C

The formula for electric potential is given by;

$\rm V=Ed$

$\rm V=Ed \\\\ \rm V=200 \times 0.05 \\\\ \rm V= 10 \frac{Nm}{C} = 10 \frac{J}{C} = 10 V.$

The work is defined as the product of the potential difference and charge of an electron.

$\rm W= 10 \times 1.602 x 10^{-19} \\\\\ \rm W= 1.6 x 10^{-18 }J$

Hence the electric potential difference and the work will be 10V and 1.6 x 10^-18 J respectively.

To learn more about the electric field refer to the link;

brainly.com/question/15071884

8 0

2 years ago

Question 1 of 4 Attempt 4 The acceleration due to gravity, ???? , is constant at sea level on the Earth's surface. However, the

Evgen [1.6K]

Answer:

g(h) = g ( 1 - 2(h/R) )

<em>*At first order on h/R*</em>

Explanation:

Hi!

We can derive this expression for distances h small compared to the earth's radius R.

In order to do this, we must expand the newton's law of universal gravitation around r=R

Remember that this law is:

$F = G \frac{m_1m_2}{r^2}$

In the present case m1 will be the mass of the earth.

Additionally, if we remember Newton's second law for the mass m2 (with m2 constant):

$F = m_2a$

Therefore, we can see that

$a(r) = G \frac{m_1}{r^2}$

With a the acceleration due to the earth's mass.

Now, the taylor series is going to be (at first order in h/R):

$a(R+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R}$

a(R) is actually the constant acceleration at sea level

and

$a(R) =G \frac{m_1}{R^2} \\ \frac{da(r)}{dr}_{r=R} = -2 G\frac{m_1}{R^3}$

Therefore:

$a(R+h) \approx G\frac{m_1}{R^2} -2G\frac{m_1}{R^2} \frac{h}{R} = g(1-2\frac{h}{R})$

Consider that the error in this expresion is quadratic in (h/R), and to consider quadratic correctiosn you must expand the taylor series to the next power:

$a(R+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R} + \frac{h^2}{2!} \frac{d^2a(r)}{dr^2}_{r=R}$

6 0

3 years ago

Read 2 more answers