Ask question

Ask question

All categories

MA_775_DIABLO [31]

3 years ago

14

What is the ratio of the gravitational force on an astronaut when he is in a space satellite in a circular orbit 315 km above th

e earth to the gravitational force on the astronaut when he is on the surface of the earth? The earth’s diameter is 12760 km.

1 answer:

Nina [5.8K]3 years ago

8 0

Answer:

The ratio of gravitational force is 0.8920

Explanation:

The gravitational force is given by F= $\frac{GMm}{R^{2} }$

The gravitational force on the astronaut when he is in

Surface of earth F₁ = $G\frac{Mm}{R^{2} }$ eqn 1

Circular orbit F₂ = $G\frac{Mm}{(R+h)^{2} }$ eqn 2

Where M is mass of earth,

m is mass of astronaut

R is radius of earth 6380 km,

h is the distance from earth surface 315 km.

Dividing eqn 1 and 2

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

= $\frac{(6380)^{2} }{(6380+315)^{2} }$

=0.8920

The ratio of gravitational force is 0.8920

You might be interested in

A ball is dropped from rest from the top of a building of height h. At the same instant, a second ball is projected vertically u

uranmaximum [27]

Answer:

a) $t = \sqrt{\frac{h}{2g}}$

b) Ball 1 has a greater speed than ball 2 when they are passing.

c) The height is the same for both balls = 3h/4.

Explanation:

a) We can find the time when the two balls meet by equating the distances as follows:

$y = y_{0_{1}} + v_{0_{1}}t - \frac{1}{2}gt^{2}$

Where:

$y_{0_{1}}$ : is the initial height = h

$v_{0_{1}}$ : is the initial speed of ball 1 = 0 (it is dropped from rest)

$y = h - \frac{1}{2}gt^{2}$ (1)

Now, for ball 2 we have:

$y = y_{0_{2}} + v_{0_{2}}t - \frac{1}{2}gt^{2}$

Where:

$y_{0_{2}}$ : is the initial height of ball 2 = 0

$y = v_{0_{2}}t - \frac{1}{2}gt^{2}$ (2)

By equating equation (1) and (2) we have:

$h - \frac{1}{2}gt^{2} = v_{0_{2}}t - \frac{1}{2}gt^{2}$

$t=\frac{h}{v_{0_{2}}}$

Where the initial velocity of the ball 2 is:

$v_{f_{2}}^{2} = v_{0_{2}}^{2} - 2gh$

Since $v_{f_{2}}^{2}$ = 0 at the maximum height (h):

$v_{0_{2}} = \sqrt{2gh}$

Hence, the time when they pass each other is:

$t = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}}$

b) When they are passing the speed of each one is:

For ball 1:

$v_{f_{1}} = - gt = -g*\sqrt{\frac{h}{2g}} = - 0.71\sqrt{gh}$

The minus sign is because ball 1 is going down.

For ball 2:

$v_{f_{2}} = v_{0_{2}} - gt = \sqrt{2gh} - g*\sqrt{\frac{h}{2g}} = (\sqrt{1} - \frac{1}{\sqrt{2}})*\sqrt{gh} = 0.41\sqrt{gh}$

Therefore, taking the magnitude of ball 1 we can see that it has a greater speed than ball 2 when they are passing.

c) The height of the ball is:

For ball 1:

$y_{1} = h - \frac{1}{2}gt^{2} = h - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h$

For ball 2:

$y_{2} = v_{0_{2}}t - \frac{1}{2}gt^{2} = \sqrt{2gh}*\sqrt{\frac{h}{2g}} - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h$

Then, when they are passing the height is the same for both = 3h/4.

I hope it helps you!

7 0

3 years ago

A 12.0-cm long cylindrical rod has a uniform cross-sectional area A = 5.00 cm2. However, its density increases linearly from 2.6

andriy [413]

Answer:

(a) The constants required describing the rod's density are B=2.6 and C=1.325.

(b) The mass of the road can be found using $A\int_0^{12}\left(B+Cx)dx$

Explanation:

(a) Since the density variation is linear and the coordinate x begins at the low-density end of the rod, we have a density given by

$2.6\frac{g}{cm^3}+\frac{18.5\frac{g}{cm^3}-2.6\frac{g}{cm^3}}{12 cm}x = 2.6\frac{g}{cm^3}+1.325x\frac{g}{cm^2}$

recalling that the coordinate x is measured in centimeters.

(b) The mass of the rod can be found by having into account the density, which is x-dependent, and the volume differential for the rod:

$m=\int\rho dv=\int\left(B+Cx\right)Adx=5\int_0^{12}\left(2.6+1.325x\right)dx=126.6$ ,

hence, the mass of the rod is 126.6 g.

7 0

3 years ago

Select the correct answer.

Gelneren [198K]

Answer: It's b and c I got it right

Explanation:

Hope this helped!!!! :)

8 0

3 years ago

A magnesium oxide component must not fail when a tensile stress of 10.5 MPa is applied. Determine the maximum allowable surface

Aloiza [94]

Answer:

Maximum permitted surface crack length is 1.29 mm

Explanation:

As per the question:

Tensile stress, $\sigma = 10.5\ MPa = 10.5\times 10^{6}\ Pa$

Surface energy of magnesium oxide, SE = $1.0\ J/m^{2}$

Modulus of elasticity of the material, E = 225 GPa = $225\times 10^{9}\ Pa$

Now,

To calculate the maximum allowable surface crack length:

$L = \frac{2E\times SE}{\sigma^{2}\pi }$

$L = \frac{2\times 225\times 10^{9}\times 1.0}{10.5\times 10^{6}\times \pi } = 1.29\times 10^{- 3}\ m = 1.29\ mm$

7 0

3 years ago

Kiley went 5.7 km/h north and then went 5.8 km/h west. From start to finish, she went 8.1 km/h northwest.

avanturin [10]

5.7 km/h north and 5.8 km/h west are instantaneous velocities, while 8.1 km/h northwest is the average velocity.<span>

The answer choice above is correct.

The instantaneous velocities are the actual </span>velocities while traveling ( the velocity during that instant ). The average velocity is the average of the instantaneous velocities ( the speed in one direction equivalent to the two speeds <span>in different directions ).

I used speed in the explanation because velocity is speed with direction.</span>

7 0

3 years ago

Read 2 more answers