Question

Vanus has a diameter of 12100 km and a mass of 4.87*10^24 kg. Calculate the energy needed to lift one kilogram from the surface of venus to a space station in orbit 900km from the surface.

I believe we should use the Gravitational potential which is
phi = -GM\r .
But I dunnow im not getting the right answer . which is

6.95 × 106 J

Answer 1

a at surface = 8.879612267 m/s^2
u at surface = mah = 53721654.22 J
a at space station =6.728761617 m/s^2
u at space station = 46764893.24 J
delta u =6956760.98 J
by using potential energy

Answer 2

\[a = \frac{ M G }{ d ^{2} }\]

Answer 3

oh that means on the surface on venus and close to the space station has diff g ?

Answer 4

Ok The force on the object is
\[F=\frac{ GMm }{ r ^{2} }\]

the body moves from position r = radius of planet to r= radius of orbit

Work done = Fx distance, although in this case because force id not constant

\[Work Done =\int\limits_{r _{p}}^{r _{o}}\ \frac{ GMm }{ r ^{2} }dr\]

Answer 5

since GMm is a constant this becomes
\[Energy required = work done = GMm \int\limits_{r _{p}}^{r _{o}}\frac{ 1 }{ r ^{2} }dr\]

Answer 6

Note - this only deals with the increase in potential energy.

I have assumed that this is what is required. However - to reach an 'orbiting' space station the mas ALSO needs to acquire an large increase in Kinetic energy (to reach the orbital velocity of the space staition)

IF your current study subject is including orbital velocity tehn you need to consider this in addition to the increase in Potential Energy.....

Answer 7

Thank you Mr. Nood :)

Answer 8

@Lama97 acceleration differs
but for small distance on the Earth it is unmeasurable ,so books just say it is constant (if you calculated it it won't affect much in small distances)

Answer 9

@catch.me
That is not correct for this question

the value of g (approx. 9.81m/s^2) is the value og GM/r^2 for a body on earth.

radius of earth is approx. 6400km

Since the force is proportional to 1/r^2 it only takes about a 5% change in r to make a 10% change in force so you can only use g as a constant ON earth and CLOSE to surface.

In THIS case the planet is not earth, and the change in radius from surface to orbit is about 8%.

The integration method is required to get accurate results in this case and similar...

Answer 10

@MrNood Hey man i calculated a1 at surface and a2 at the orbit by a= MG/r^2
then used the formula for potential energy u= mah.
and subtracted them why would i integrate it is simple problem ??

Answer 11

@catch.me
It is not that simple

Work done in simple terms = force x distance

But that applies for a constant force. For a varying force the work done over a small distance delta h is given by f x delta h

What you have done is taken the start and end points - BUT the force varies continuously, getting progressively weaker as h increase.

Thus f is a function of h

so at any point as you raise the weight the force acting is GMm/(h +rp)^2

and the work done to raise another small distance AT THAT POINT is f * delta h

The only way to sum the total effect of the force over the whole range is by integration where delta h become dh

(note that we have used h in your argument - it is comparable to r in my expression)

Answer 12

r in my expression is distance from centre of planet

So if you use h (for height above surface you have to add rp (i.e. radius of planet) as I did above to convert h to r

Answer 13

By the way:

Your formula for potential energy is comparable to the 'f x d' formula I gave above.

For a body on earth ma = mg = weight, and h is height

so ma x h = force x distance

BUT BUT BUT
this goes back to the original simplification that a=g= constant.

That is perfectly good assumption for changes in h which are SMALL compared to the radius of the planet - but is NOT valid for this question and other 'orbit' type quedtions.

Answer 14

Yea that is true
i forgot that i have to put the main formula for potential energy like acceleration.
@Lama97 follow @MrNood