Give reason ?

I'll include a calculation just to prove that,

#g=(Gm)/r^2#, where:

• #g# = acceleration due to free fall (#ms^-2#)
• #G# = gravitational constant (#6.67*10^-11Nm^2# #kg^-2#)
• #m# = mass of object (#kg#)
• #r# = distance between a point and center of mass of the object (#m#)

Since #G# is a constant, we can say that:
#(g_Er_E""^2)/(m_E)=(g_Jr_J""^2)/(m_J)#

#(9.8r_E""^2)/(m_E)=(25(ar_E)^2)/(319m_E)#

#9.8=(25a^2)/319#

#a^2=(9.8*319)/25#

#a=sqrt((9.8*319)/25)~~11.2#

Radius of Jupiter is approximately 11.2 times that of Earth.

An increase in mass results in an increase in acceleration due to free fall, but an increase in distance decreases acceleration. Since #DeltaM>Deltar^2#, acceleration will be greater.

We define acceleration due to gravity at the surface of the earth as

#g-=GM_e/R_e^2# .........(1)
where #G# is Universal Gravitational Constant, #M_eand R_e# are mass and radius of earth respectively. After inserting varius values we have #g=9.81\ ms^-2#

Similarly acceleration due to gravity at the surface of Jupiter would be

#g_j=(GM_j)/R_j^2# .......(2)

Dividing (2) by (1) and inserting given values we get

#g_j/g=M_j/M_e R_e^2/R_j^2#
#=>25/9.8=319xxR_e^2/R_j^2#
#=>R_e/R_j=sqrt(25/9.8xx1/319)#
#=>R_e/R_j=0.089#

Radius of earth #=6,371\ km#
Radius of Jupiter #=69,911\ km#

#=>R_e/R_j=6371/69911=0.091#

We see that both ratios match well.

Therefore, we can say that gravity of Jupiter is only #2.5times# that of earth due to larger radius of Jupiter which appears in the inverse square expression of gravity.