# A-level Physics (Advancing Physics)/Gravitational Fields/Worked Solutions

G = 6.67 x 10−11 m3kg−1s−2

1. A 15 kg object has a weight of 8000N. What is the gravitational field strength at this point?

${\displaystyle g={\frac {F_{grav}}{m}}={\frac {8000}{15}}=533{\mbox{ Nkg}}^{-1}}$

2. Draw a graph of gravitational field strength against distance.

Since gravitational field strength is proportional to gravitational force, the two graphs look very similar.

3. What is the gravitational field strength of the Sun (mass 2 x 1030kg) on the Earth (mass 6 x 1024kg, mean orbital radius 15 x 1010m)?

${\displaystyle g={\frac {GM}{r^{2}}}={\frac {6.67\times 10^{-11}\times 2\times 10^{30}}{(15\times 10^{10})^{2}}}=5.93{\mbox{ mNkg}}^{-1}}$

4. What is the difference in the acceleration due to gravity over a vertical distance d?

${\displaystyle \Delta g={\frac {GM}{(r+d)^{2}}}-{\frac {GM}{r^{2}}}=GM\left({\frac {1}{(r+d)^{2}}}-{\frac {1}{r^{2}}}\right)={\frac {GM(r^{2}-(r+d)^{2})}{r^{2}(r+d)^{2}}}={\frac {GM(r^{2}-r^{2}-2rd-d^{2})}{r^{2}(r+d)^{2}}}={\frac {-GMd(2r+d)}{r^{2}(r+d)^{2}}}}$

5. How far would one have to travel upwards from the Earth's surface to notice a 1Nkg−1 difference in gravitational field? (The Earth has a radius of 6400 km.)

${\displaystyle 1={\frac {GMd(2r+d)}{r^{2}(r+d)^{2}}}}$

${\displaystyle GMd(2r+d)=r^{2}(r+d)^{2}}$

${\displaystyle 2GMrd+GMd^{2}=r^{2}(r^{2}+2rd+d^{2})=r^{4}+2r^{3}d+r^{2}d^{2}}$

${\displaystyle d^{2}(r^{2}-GM)+dr(2r^{2}-2GM)+r^{4}=0}$

${\displaystyle d={\frac {2GMr-2r^{3}\pm {\sqrt {r^{2}(2r^{2}-2GM)^{2}-4r^{4}(r^{2}-GM)}}}{2(r^{2}-GM)}}={\frac {2GMr-2r^{3}\pm {\sqrt {4r^{6}-8GMr^{4}+4G^{2}M^{2}r^{2}-4r^{6}+4GMr^{4}}}}{2(r^{2}-GM)}}}$

${\displaystyle ={\frac {2GMr-2r^{3}\pm r{\sqrt {4r^{4}-8GMr^{2}+4G^{2}M^{2}-4r^{4}+4GMr^{2}}}}{2(r^{2}-GM)}}}$

Which is very horrible. If you plug in the numbers, you get:

${\displaystyle d={\frac {4.598272\times 10^{2}1\pm 4.853341552\times 10^{21}}{-7.1848\times 10^{14}}}}$

Since we want a negative d (because of the minus sign we ignored right at the beginning):

${\displaystyle d={\frac {4.598272\times 10^{2}1+4.853341552\times 10^{21}}{-7.1848\times 10^{14}}}=-13155{\mbox{ km}}}$

Although, I have to confess, there was so much scope for error that it is almost certain that this is the wrong answer. If you fancy wading through and checking, feel free!

[*] Solution :

g'=g-1=9.8-1=8.8N/kg g=GM/(R+d)^2 => ${\displaystyle (R+d)^{2}=GM/g'=>R+d=sqrt(GM/g')=>d=sqrt(GM/g')-R}$
d=349.96 km

--Sjlegg (talk) 14:18, 23 April 2009 (UTC)